Notes

(Corrected from the printed mark scheme: in the alternative lines the scheme prints \(\dfrac{1}{4}(k+1)^2(k+1)^2\) in place of \(\dfrac{1}{4}(k+1)^2(k+2)^2\), here and in the notes below; \(\sum_{r=1}^{k+1} r^3 = \dfrac{1}{4}(k+1)^2(k+2)^2\).)

B1: Shows the statement is true for \(n = 1\). Minimum requirement \(1 = \dfrac{1}{4}(1)(4)\)

M1: Makes the inductive assumption, assume true \(\boldsymbol{n = k}\). This may appear in the conclusion.

M1: A correct statement for \(n = k + 1\)

M1: Factorises out \((k+1)^2\) or multiples out \(\dfrac{1}{4}k^2(k+1)^2 + (k+1)^3\) and \(\dfrac{1}{4}(k+1)^2(k+2)^2\) to form a quartic for each

A1: Achieves \(\dfrac{1}{4}(k+1)^2\left(\{k+1\} + 1\right)^2\) with no omissions or errors. Shows both are sides are \(k^4 + 6k^3 + 13k^2 + 12k + 4\) no omissions or errors. and draws a minimal conclusion. Shows \(\dfrac{1}{4}(k+1)^2(k+2)^2 - \dfrac{1}{4}k^2(k+1)^2 = (k+1)^3\) may expand both sides and draw a minimal conclusion or factorises, no omissions or errors.

Note: they may state aiming for \(\dfrac{1}{4}(k+1)^2(k+2)^2\) and then achieves this which is fine for A1

A1: Makes appropriate concluding sentence covering the points indicated in scheme. Dependent on all previous marks except B1. The statement true for \(n = 1\) could appear at the start of the solution

Notes

B1: Shows that the result holds for \(n = 1\). Must see substitution in the RHS minimum required is \(\begin{pmatrix}1 & 5(2 - 1)\\ 0 & 2\end{pmatrix}\) and reaches \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}\) No need to state “true for \(n = 1\)” for this mark.

M1: Assumes the result is true for some value of \(n = k\), and sets up a matrix multiplication of the assumed result multiplied by the original matrix, either way round (no need to carry out for this mark, it is for essentially explaining the procedure). “Assume (true for) \(n = k\)” (oe) is sufficient for the assumption and may even be part of the conclusion and accept any alternative wording that indicates the assumption has been made. Allow for setting up the multiplication in reverse (ie working from \(n = k + 1\) towards \(n = k\), \(\mathbf{A}^k = \mathbf{A}^{k+1}\mathbf{A}^{-1}\) oe)

M1: Carries out the multiplication. (Allow the reverse case.)

A1: Achieves a correct un-simplified matrix. Accept in working in reverse.

A1: Reaches a correct simplified matrix with no errors, the correct un-simplified matrix seen previously and at least one intermediate line which must be correct. If working from both sides all steps in showing the two sides are equal must be seen. If working in reverse they must return to \(\mathbf{A}^{k+1} = \ldots\) for this mark.

A1: Correct formal conclusion. This mark is dependent on the M and previous A marks having been scored and an attempt at the check for \(n = 1\) (if e.g. they didn’t show sufficient detail). It is gained by conveying the ideas of all three bold points at the end of their solution.

Note: Some cases may use \(n = 0\) as the base case. These can score full marks if dealt with correctly, but the conclusion must be consistent with their initial check to score the final A.
For the B mark minimum \(\begin{pmatrix}1 & 5\\ 0 & 2\end{pmatrix}^0 = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) ( or \(\mathbf{I}\)) and \(\begin{pmatrix}1 & 5(1 - 1)\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) should be seen. (No need to state true for \(n = 0\) for this mark – but must have consistent conclusion for the final A as noted above). If unsure send to review.

Notes

B1: Identifies the transformation as a reflection. Accept stretch with scale factor \(-1\).

B1: Identifies reflection and the correct line of reflection. Condone phrasing such as “across the \(y\)-axis”. Must be a single transformation – B0 if they state something else also happens.
NB Allow B1B0 for answers such as “flip in the y-axis” that convey the correct transformation in imprecise language.

Notes

B1: Correct matrix

Notes

M1: A complete method to find a value for \(2^n\) (or \(n\)). Multiplies the coordinates \((27,\ 1)\) by their matrix \(\mathbf{C}\) and sets the \(x\) and \(y\) coordinates equal to reach a value for \(2^n\) or \(n\) (condone negative values for \(2^n\)). Note you may allow this for reach a value of \(a\) if \(a = 2^n\) is stated or clearly implied by their working.

M1: Uses their value of \(2^n\) or \(n\), to find the matrix \(\mathbf{A}^n\). Note substituting into \(\mathbf{C}\) is M0 without further work to find \(\mathbf{A}^n\).

A1: Correct matrix.

Notes

(i) Must be using induction to score any marks

B1: Shows that the result holds for \(n = 1\). Minimum LHS = \(\dfrac{1}{1(2)}\) RHS \(= \dfrac{1}{1 + 1}\)

M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)

M1: Attempts to add the next term and makes progress by attempting a common denominator.

A1: Achieves a correct expression in terms of \(k + 1\), with no errors seen cso. Alternatively finds \(\displaystyle\sum_{r=1}^{k+1} \frac{1}{r(r + 1)} = \frac{k + 1}{k + 2}\) and reaches the same expression using induction.

A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted

Notes

Way 1: f(\(\boldsymbol{k}\) + 1)

B1: Shows that the result holds for \(n = 1\)

M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)

M1: Attempts f (\(k\) + 1) and attempts to express in terms of f (\(k\))

A1: Achieves a correct expression in terms of f (\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted

Alternative: Way 2: f(\(\boldsymbol{k}\) + 1) − f(\(\boldsymbol{k}\))

B1: Shows that the result holds for \(n = 1\)

M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)

M1: Attempts f (\(k\) + 1) – f (\(k\)) and attempts to express in terms of f (\(k\))

A1: Achieves a correct expression for f (\(k\) + 1) in terms of f (\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted

Alternative: Way 3: f(\(\boldsymbol{k}\)) = 5\(\boldsymbol{M}\)

B1: Shows that the result holds for \(n = 1\)

M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)

M1: Attempts f(\(k\) +1) and writes in terms of 5\(M\).

A1: Achieves a correct expression for f (\(k\) + 1) in terms of \(M\) and \(2^{2k}\) or \(M\) and \(3^{2k+4}\)

A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted

Alternative: Way 4: f(\(\boldsymbol{k}\) + 1) − \(\boldsymbol{m}\)f(\(\boldsymbol{k}\))

B1: Shows that the result holds for \(n = 1\)

M1: Makes a statement that assumes the result is true for some value of \(n\), say \(k\)

M1: Attempts f (\(k\) + 1) – m f (\(k\)) and attempts to express in terms of f (\(k\))

A1: Achieves a correct expression for f (\(k\) + 1) in terms of f (\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Not dependent on B1 as long as attempted

Note conclusion may be in terms of \(\mathrm{f}(1), \mathrm{f}(k), \mathrm{f}(k + 1)\)

Notes

B1: Demonstrates the statement is true for \(n = 1\). Accept as minimum \((1)^2 = 1\) and \(\dfrac{1}{3}(3) = 1\) for the check (both sides must clearly be evaluated as 1), and for the conclusion “true for \(n = 1\)” or broken “if/when \(n = 1\) <check> (hence) true/shown/tick” – must be part of the initial check (not just in the conclusion).

M1: Assume the result for \(n = k\) and attempts to add the \((k + 1)\)^th term to the result for the sum to \(k\) terms. An assumption may be clearly made, but accept tacit assumptions. Allow if there are minor slips in the expressions if the intent is clear.

dM1: Makes progress towards proving the inductive step by either factorising the common factor \((2k + 1)\) or expanding fully. There may be variations so score for equivalent progress.

A1: Either for a correct factorised form with \((k + 1)\) as a factor or a fully correct expansion with gathered terms. Must have come from correct work from the inductive step, not backwards worked from the \(n = k + 1\) expression.

A1: Completes the inductive steps by obtaining the correct expression in terms of \((k + 1)\) with suitable intermediate step e.g. clear factorisation of \((k + 1)\) first, OR by expanding the required expression for \(n = k + 1\) to achieve equal expressions. Must have been completely correct work, no errors.

A1: Depends on the MMAA marks having been scored with at least an attempt to check one side of the \(n = 1\) case having been made. Correct complete conclusion. Must include the notions of “true for \(n = 1\)”, “true for \(n = k\) implies true for \(n = k + 1\)” and “hence true for all \(n\)” though the exact wording will vary. The conclusion should be given at the end with the exception that the “true for \(n = 1\)” may be stated with the initial check.

Note: stating “true for \(n = k\) and \(n = k + 1\)” will be A0.
Note: Accept use of \(n\) instead of \(k\) for all except the final A mark.
Note: Attempts at using summation formulae without induction score no marks.

Notes

Alternative using exponential definitions

Notes

M1: Differentiates to a correct form

A1: Correct derivative and reaches appropriate form to deduce the result is true for \(n = 1\), minimum \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \mathrm{e}^{2x}\left(\dfrac{4}{2}\sinh x + \dfrac{2}{2}\cosh x\right)\)

M1: (Makes the inductive assumption and) attempts the \((k + 1)\) th derivative from the \(k\) th derivative. Allow slips in coefficients but must be evidence of the use of the product rule. Must be of the form

dM1: Dependent on the previous method mark. Factors out the exponential and gathers the \(\sinh x\) and \(\cosh x\) terms. Accept either form shown or equivalent.

A1: Reaches the correct form from correct work. Must have the “\(k\)+1” showing. Depends on the previous two method marks.

A1: Makes appropriate concluding sentence covering the points indicated in scheme. Depends on all method marks having been scored. Must have reached at least the second line shown in the dM mark, and made an attempt at checking \(n = 1\) (though the first A mark need not have been scored if insufficient detail shown).
Maybe seen as a narrative throughout their solution.
This mark requires all necessary brackets throughout.

Alternative: using exponentials
M1: Uses the exponential definition of \(\sinh x\) and differentiates to a correct form
A1: Correct derivative and uses the exponential definitions of \(\sinh x\) and \(\cosh x\) to deduce the result is true for \(n = 1\)
M1: (Makes the inductive assumption and) attempts the \((k + 1)\) th derivative from the \(k\) th derivative. Uses the exponential definition of \(\sinh x\) first and then differentiates to a correct form
dM1: Dependent on the previous method mark. Collects exponential terms and simplifies
A1: Reaches the \(k + 1\) th derivative. Uses the exponential definitions of \(\sinh x\) and \(\cosh x\) in the result, simplifies and achieves the correct result
A1: Makes appropriate concluding sentence covering the points indicated in scheme. Depends on all method marks having been scored. Must have reached at least the second line shown in the dM mark, and made an attempt at checking \(n = 1\) (though the first A mark need not have been scored if insufficient detail shown).
Maybe seen as a narrative throughout their solution.
This mark requires all necessary brackets throughout.

Notes

B1: Shows true for \(n = 1\).
Need to see \(n = 1\) substituted into rhs. The minimum for B1 would be \(\begin{pmatrix}1 & -2 \times 1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}\).
There is no need to state “True for \(n = 1\)”

M1: (Assumes for \(n = k\) and) multiplies original matrix by \(k\)th power matrix either way round. Note that the assumption statement is not needed for this mark (but see below) so just look for:

A1: Achieves a correct unsimplified matrix.

A1: Reaches correct form for the matrix with the \(k + 1\) factored out with no errors and the correct unsimplified matrix seen previously. Note that the result may be proved by equivalence (see below).

A1: Correct conclusion with assumption made (which may be implied in their conclusion if they say “if true for \(n = k\) then…”). This mark is dependent on all the previous marks apart from the B mark and is gained by conveying all the underlined points.
Allow this mark to score as long as all the underlined points are seen as narrative in their solution.
There must be the assumption statement somewhere or the “if…then…” idea in the conclusion. If awarded for the assumption statement condone e.g. true for \(n = k\) in the conclusion.

The conclusion must convey the “if true for \(n = k\) then true for \(n = k + 1\)” idea and not e.g. true for \(k\), true for \(k + 1\), true for 1 therefore…but see the previous note.
For the “true for all \(n\)” part condone e.g. “true for \(n\)”, “true for all integers after 1”, “true for \(\mathbb{Z}^+\)”,
But do not allow “true for all values”, “true for all real numbers”

Q4 Extra Notes:

1. For candidates who use \(n\) instead of \(k\) throughout withhold the final mark if the work is otherwise correct.

2. For equivalence proofs, this would be minimally acceptable:

\(\begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2 \times 1\\ 0 & 1\end{pmatrix}\) B1

“Need to prove” oe e.g. “target is”, “\(n = k + 1\)”, “need” etc. \(\begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}^{k+1} = \begin{pmatrix}1 & -2k - 2\\ 0 & 1\end{pmatrix}\)

Note that there must be a reference to \(k + 1\) as above or e.g. (Target =) \(\begin{pmatrix}1 & -2(k + 1)\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2k - 2\\ 0 & 1\end{pmatrix}\)

\(\begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}^{k+1} = \begin{pmatrix}1 & -2\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & -2k\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & -2k - 2\\ 0 & 1\end{pmatrix}\) M1A1A1

Hence result is true for \(n = k + 1\). As true for \(n = 1\) and have shown if true for \(n = k\) then it is true for \(n = k + 1\), so it is true for all \(n\). A1

Without the “Need to prove…” oe the response would score B1M1A1 and then A1 if the factorised form was shown or equivalence shown and then A1 for the correct conclusion.

3. Allow e.g. “correct” for “true” in the conclusion

Notes

B1: Shows the statement is true for \(n = 1\). Accept as minimum \(\begin{pmatrix}1 - 6 & 9\\ -4 & 1 + 6\end{pmatrix} = \begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\)

M1: Makes the inductive assumption, assume true \(\boldsymbol{n = k}\). This may appear in the conclusion.

M1: A correct statement for \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^{k+1}\) in terms of \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}^k\), can be either way round.
Can be implied by \(\begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\times\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\) or \(\begin{pmatrix}-5 & 9\\ -4 & 7\end{pmatrix}\times\begin{pmatrix}1 - 6k & 9k\\ -4k & 1 + 6k\end{pmatrix}\)

M1: Carries out the multiplication correctly, condone sign slips

A1: Correct simplified matrix from fully correct working

A1: Completes the inductive argument by showing clearly the matrix has the correct form (must have \((k + 1)\) factors in terms) or uses the result with \(n = k + 1\) and shows that their result is the same.
Conclusion conveying all three underlined points or equivalent at some point in their argument. Depends on all three M’s and A marks but can be scored without the B mark as long as it is stated true for \(n = 1\)

Notes

B1: Shows that the result holds for \(n = 1\). Must see substitution in the RHS minimum required is \(\begin{pmatrix}3 & \frac{a}{2}(3 - 1)\\ 0 & 1\end{pmatrix}\) and reaches \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}\)

M1: Assumes the result is true for some value of \(n = k\). Assume (true for) \(n = k\) is sufficient. Alternatively states assume \(\mathbf{M}^n\) or \(\begin{pmatrix}3 & a\\ 0 & 1\end{pmatrix}^k = \begin{pmatrix}3^{k} & \dfrac{a}{2}\left(3^{k} - 1\right)\\ 0 & 1\end{pmatrix}\)

M1: Sets up a matrix multiplication of their assumed result multiplied by the original matrix, either way round. Allow a slip as long as the intention is clear.

A1: Achieves a correct un-simplified matrix

A1: Reaches a correct simplified matrix with no errors, the correct un-simplified matrix seen previously and at least one intermediate line which must be correct.

A1: Correct conclusion. This mark is dependent on all previous marks except B mark but \(n = 1\) must have been attempted. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution. Condone \(n \in \mathbb{Z}\)

Notes

(i) B1: States correct determinant. This can be implied by a correct equation

M1: Correct method to find a value of \(n\) using \(5 \times \text{`their } \det\left(\mathbf{M}^n\right)\text{'} = 1215\) which involves solving an index equation of the form \(p^n = q\) where \(n \gt 1\)

A1: \(n = 5\)

(ii) M1: Sets up an equation by multiplying the matrix \(\mathbf{M}^n\) by \(\begin{pmatrix}2\\ -2\end{pmatrix}\) setting equal to \(\begin{pmatrix}123\\ -2\end{pmatrix}\) and reaches a value for \(a\). You may just see \(2\left(3^n\right) - 2\dfrac{a}{2}\left(3^n - 1\right) = 123 \Rightarrow a = \ldots\)
Follow through on their value for \(n\).

A1: \(a = 1.5\)

(Corrected from the printed mark scheme: in the inverse-matrix line the printed scheme has \(a\) in place of 243 in the bottom right of the inverse matrix, and the equation as \(\dfrac{123 - 2\frac{a}{2}(243 - 1)}{243} = -2\).)

Notes

Way 1: \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)

B1: Shows that f(1) = 35 and concludes or shows divisible by 7. This may be seen in the final statement.

M1: Makes a statement that assumes the result is true for some value of \(n\)

M1: Attempts \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)

A1: Achieves a correct expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) in terms of \(\mathrm{f}(k)\)

A1: Reaches a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)

A1: Correct conclusion. This mark is dependent on all previous marks, look out for scoring B1 in this statement. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution.

Alternative: Way 2: \(\mathrm{f}(k + 1)\)

B1: Shows that f(1) = 35 and concludes divisible by 7

M1: Makes a statement that assumes the result is true for some value of \(n\)

M1: Attempts \(\mathrm{f}(k + 1)\)

A1: Correctly obtains either \(2\mathrm{f}(k)\) or \(7 \times 3^{2k+1}\) or either \(9\mathrm{f}(k)\) or \(-7 \times 2^{k+2}\)

A1: Reaches a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)

A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution.

Alternative: Way 3: \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\)

B1: Shows that f(1) = 35 and concludes divisible by 7

M1: Makes a statement that assumes the result is true for some value of \(n\)

M1: Attempts \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\)

A1: Achieves a correct expression for \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\) in terms of \(\mathrm{f}(k)\)

A1: Reaches a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)

A1: Correct conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all four bold points either at the end of their solution or as a narrative in their solution.

Notes

(i)
B1: Shows the statement is true for \(n = 1\) by evaluating both sides. There is no need for statement “hence true for \(n = 1\)” for this mark but if they never state this the final A will be forfeited.
Look for a minimum of \(4 \times 3 = 12\) for the LHS and \(1 \times 3 \times 4\) for the RHS. If only one side is explicitly evaluated, it is B0, but all other marks may be gained.
M1: Makes an assumption statement that assumes the result is true for \(n = k\)
M1: Makes the inductive step by attempting to add the \((k + 1)\)^th term to the assumed result. Attempts at using the standard summation formulae score M0, as the question requires induction.
A1: Correct expression with at least one correct linear factor
A1: Obtains a fully correct factorised expression. May be as in scheme or in terms of \(k + 1\).
A1: Correct complete conclusion with all ideas conveyed at the end or as a narrative and the sum to \(k+1\) expressed in terms of \(k + 1\) (or with the expression in term of \(k + 1\) stated earlier – it must be seen at some stage). Allow slips in notation if the intent is correct. Depends on all except the B mark, though an attempt at checking the \(n = 1\) case must have been made.

Notes

(ii) Way 1
B1: Shows that \(\mathrm{f}(1) = 15\)
M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Attempts \(\mathrm{f}(k + 2)\)
A1: Correctly obtains \(16\mathrm{f}(k)\) or \(45 \times 5^{k-1} + 120 \times 6^{k-1}\)
A1: Reaches a correct expression for \(\mathrm{f}(k + 2)\) in terms of \(\mathrm{f}(k)\)
A1: Correct conclusion, including explanation that all terms on RHS are divisible by 15, or writing terms each explicitly as multiples of 15. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution. Do not allow “true for all \(n\)” where \(n\) represents natural numbers, as this is incorrect.

Way 2
B1: Shows that \(\mathrm{f}(1) = 15\)
M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Attempts \(\mathrm{f}(k + 2) - \mathrm{f}(k)\) or equivalent work
A1: Achieves a correct expression for \(\mathrm{f}(k + 2) - \mathrm{f}(k)\) in terms of \(\mathrm{f}(k)\)
A1: Reaches a correct expression for \(\mathrm{f}(k + 2)\) in terms of \(\mathrm{f}(k)\)
A1: Correct conclusion, including explanation that all terms on RHS are divisible by 15, or writing terms each explicitly as multiples of 15. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution. Do not allow “true for all \(n\)” where \(n\) represents natural numbers, as this is incorrect.

Way 3
B1: Shows that \(\mathrm{f}(1) = 15\)
M1: Makes a statement that assumes the result is true for some odd value of \(n\) (Assume (true for) \(n = 2k + 1\) is sufficient (or may use \(2k - 1\) – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)
M1: Attempts \(\mathrm{f}(2k+3)\) (like Way 1), or \(\mathrm{f}(2k + 3) - \mathrm{f}(2k+1)\) or equivalent work (like Way 2) It must be the correct \(\mathrm{f}(2k+1)\) used, in the latter case (subtracting \(4^k + 5^k + 6^k\) instead of \(4^{2k+1} + 5^{2k+1} + 6^{2k+1}\) is M0)
A1: Reaches \(16\left(4^{2k+1} + 5^{2k+1} + 6^{2k+1}\right) + 9 \times 5^{2k+1} + 20 \times 6^{2k+1}\) or suitable equivalent OR achieves a correct expression for \(\mathrm{f}(2k + 3) - \mathrm{f}(2k+1)\) in terms of \(\mathrm{f}(2k+1)\) where factors of 15 are apparent.
A1: Reaches \(16\left(4^{2k+1} + 5^{2k+1} + 6^{2k+1}\right) + 45 \times 5^{2k} + 120 \times 6^{2k}\) or a suitable equivalent OR a correct expression for \(\mathrm{f}(2k + 3)\) of form \(A\mathrm{f}(2k + 1) + 15(\ldots)\) (though the \(15(\ldots)\) could be separate multiples of 15 where the 15 need not yet be extracted).
A1: Correct conclusion, including explanation that all terms on RHS are divisible by 15, or writing terms each explicitly as multiples of 15. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution. Do not allow “true for all \(n\)” where \(n\) represents natural numbers, as this is incorrect, but if \(n\) represents odd numbers than allow.

Note: \(4^{2k} = 16^k\) etc may be used throughout.
Note: Way 3 should apply to cases where the question is rephrased in terms of \(4^{2n+1} + 5^{2n+1} + 6^{2n+1}\) for all \(n\) (rather than odd \(n\)).
Accept use of alternative equivalent language throughout.

(corrected from the printed mark scheme: in the Way 1 conclusion, “so 15 divides \(\mathrm{f}(k+2)\)” is printed as “so 15 divides \(\mathrm{f}(k+1)\)”)

Notes

Way 1 \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)

B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)

M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc

M1: Attempts \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) or equivalent work

A1: Achieves a correct simplified expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\)

A1: Achieves a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)

A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 2 \(\mathrm{f}(k + 1)\)

B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)

M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc

M1: Attempts \(\mathrm{f}(k + 1)\)

A1: Correctly achieves either \(9\mathrm{f}(k)\) or \(5 \times 2^{2k}\) or either \(4\mathrm{f}(k)\) or \(5 \times 3^{2k+4}\)

A1: Achieves a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(\mathrm{f}(k)\)

A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 3 \(\mathrm{f}(k) = 5M\)

(Corrected from the printed mark scheme: the first line is printed as \(3^2 \times \left(5M + 2^{2k+2}\right)\); since \(3^{2k+4} = 5M + 2^{2k}\), it should be \(3^2 \times \left(5M + 2^{2k}\right)\).)

B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)

M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc

M1: Attempts \(\mathrm{f}(k + 1)\)

A1: Correctly achieves either \(45M\) or \(5 \times 2^{2k}\) or either \(20M\) or \(5 \times 3^{2k+4}\)

A1: Achieves a correct expression for \(\mathrm{f}(k + 1)\) in terms of \(M\) and \(2^{2k}\) or \(M\) and \(3^{2k+4}\)

A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 4 \(\mathrm{f}(k + 1) + \mathrm{f}(k)\)

B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)

M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc

M1: Attempts \(\mathrm{f}(k + 1) + \mathrm{f}(k)\) or equivalent work

A1: Achieves a correct simplified expression for \(\mathrm{f}(k + 1) + \mathrm{f}(k)\)

A1: Achieves a correct expression for \(\mathrm{f}(k + 1) = 5\left[2 \times 3^{2k+4} - 2^{2k}\right] - \mathrm{f}(k)\)

A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 5 \(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k)\)

(Selecting a value of M that will lead to multiples of 5)

Way 5 \(\mathrm{f}(k + 1) - M\mathrm{f}(k)\) (Selects a suitable value for M which leads to divisibility of 5)

B1: Shows the statement is true for \(n = 1\). Needs to show \(\mathrm{f}(1) = 725\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)

M1: Makes an assumption statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc

M1: Attempts \(\mathrm{f}(k + 1) - M\mathrm{f}(k)\) or equivalent work

A1: Achieves a correct simplified expression, \(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k)\) which is divisible by 5

\(\mathrm{f}(k + 1) - \text{‘}M\text{’}\mathrm{f}(k) = \left(9 - \text{‘}M\text{’}\right) \times 3^{2k+4} - \left(4 - \text{‘}M\text{’}\right) \times 2^{2k}\)

A1: Achieves a correct expression for \(\mathrm{f}(k + 1) = \left(9 - \text{‘}M\text{’}\right) \times 3^{2k+4} - \left(4 - \text{‘}M\text{’}\right) \times 2^{2k} + \text{‘}M\text{’}\mathrm{f}(k)\) which is divisible by 5

A1: Correct complete conclusion. This mark is dependent on all previous marks. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.

Notes

B1: Substitutes \(n = 1\) into both sides of the statement to show they are equal. As a minimum expect to see \(\dfrac{1}{1 \times 3}\) and \(\dfrac{1}{2 + 1}\) for the substitutions. (No need to state true for \(n = 1\) for this mark.)

M1: Assumes (general result) true for \(n = k\). (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then ... etc.)

M1: Attempts to add \((k + 1)\)th term to their sum of \(k\) terms. Must be adding the \((k + 1)\)th term but allow slips with the sum.

dM1: Depends on previous M. Combines their two fractions over a correct common denominator for their fractions, which may be \((2k + 1)^2(2k + 3)\) (allow a slip in the numerator).

A1: Correct algebraic work leading to \(\dfrac{(k + 1)}{2(k + 1) + 1}\) or \(\dfrac{k + 1}{2k + 3}\)

A1: cso Depends on all except the B mark being scored (but must have an attempt to show the \(n = 1\) case). Demonstrates the expression is the correct for \(n = k + 1\) (both sides must have been seen somewhere) and gives a correct induction statement with all three underlined statements (or equivalents) seen at some stage during their solution (so true for \(n = 1\) may be seen at the start).

For demonstrating the correct expression, accept giving in the form \(\dfrac{(k + 1)}{2(k + 1) + 1}\), or reaching \(\dfrac{k + 1}{2k + 3}\) and stating “which is the correct form with \(n = k + 1\)” or similar – but some indication is needed.

Note: if mixed variables are used in working (\(r\)’s and \(k\)’s mixed up) then withhold the final A.

Note: If \(n\) is used throughout instead of \(k\) allow all marks if earned.

Notes

B1: Shows that the result holds for \(n = 1\). Must see substitution into the rhs.
The minimum would be: \(\begin{pmatrix}4 + 1 & -8\\ 2 & 1 - 4\end{pmatrix} = \begin{pmatrix}5 & -8\\ 2 & -3\end{pmatrix}\).

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Sets up a correct multiplication statement either way round

A1: Achieves a correct un-simplified matrix

A1: Reaches a correct simplified matrix with no errors and the correct un-simplified matrix seen previously. Note that the simplified result may be proved by equivalence.

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Notes

(ii) Way 1

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1) – f(\(k\)) or equivalent work

A1: Achieves a correct expression for f(\(k\) + 1) – f(\(k\)) in terms of f(\(k\))

A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 2: \(\mathrm{f}(k + 1)\)

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1)

A1: Correctly obtains 4f(\(k\)) or \(21 \times 5^{2k-1}\)

A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 3: \(\mathrm{f}(k + 1) - m\mathrm{f}(k)\)

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1) – \(m\)f(\(k\))

A1: Achieves a correct expression for f(\(k\) + 1) – \(m\)f(\(k\)) in terms of f(\(k\))

A1: Reaches a correct expression for f(\(k\) + 1) in terms of f(\(k\))

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Alternative: Way 4: \(\mathrm{f}(k) = 21M\)

B1: Shows that f(1) = 21

M1: Makes a statement that assumes the result is true for some value of \(n\) (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then … etc.)

M1: Attempts f(\(k\) + 1)

A1: Correctly obtains \(84M\) or \(21 \times 5^{2k-1}\)

A1: Reaches a correct expression for f(\(k\) + 1) in terms of \(M\) and \(5^{2k-1}\)

A1: Correct conclusion. This mark is dependent on all previous marks apart from the B mark. It is gained by conveying the ideas of all four underlined points either at the end of their solution or as a narrative in their solution.

Typical solution

(i)

(ii)

so \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\)

Typical solution

When \(n = 1\):

so, the rule is true for \(n = 1\)

Assume the rule is true for \(n = k\)

so, the rule is also true for \(n = k + 1\)

So, by induction,

is true for all \(n \in \mathbf{N}\)

Typical solution

(i)

Astrid did not show that the result is true for \(n = 1\)

(ii)

and

Typical solution

(i)

It is true for \(n = k\), and also true for \(n = k + 1\)

(ii)

If true for \(n = k\), it is also true for \(n = k + 1\)

Typical solution

3 is divisible by 3

So the rule is true for \(n = 1\)

If the rule is true for \(n = k\) then \(5^k - 2^k = 3m\) for some integer \(m\)

So the rule is also true for \(n = k + 1\)

Therefore, by induction

\(5^n - 2^n\) is divisible by 3 for all \(n \in \mathbf{N}\)

Typical solution

Let \(n = 1\); then the formula gives

(So the formula is true for \(n = 1\))

Assume the formula is true for \(n = k\)

Then we want to show that

But

as required

So the formula is also true for \(n = k + 1\)

The formula for \(u_n\) is true for \(n = 1\)

If the formula is true for \(n = k\), then the formula is also true for \(n = k + 1\)

Hence by induction \(u_n = \dfrac{3}{2}n^2 - \dfrac{3}{2}n + 1\) is true for all integers \(n \geqslant 1\)

Typical solution

When \(n = 1\):

\(\therefore\) the rule is true for \(n = 1\)

Assume it is true for \(n = k\)

\(\therefore\) the rule is also true for \(n = k + 1\)

So, by induction,

is true for all integers \(n \geqslant 1\)

Typical solution

Typical solution

Typical solution

Typical solution

Let \(n = 1\); then the formula gives

so the result is true for \(n = 1\)

Assume the result is true for \(n = k\), so

Then

and the result also holds for \(n = k + 1\)

\(\mathrm{f}(n)\) is divisible by 19 for \(n = 1\); if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction \(\mathrm{f}(n)\) is divisible by 19 for \(n \geqslant 1\)

Typical solution

Let \(n = 1\): \((\mathbf{ABA}^{-1})^1 = \mathbf{ABA}^{-1} = \mathbf{AB}^1\mathbf{A}^{-1}\)

\(\therefore\) it is true for \(n = 1\)

If it is true for \(n = k\), then

Therefore, by induction,

is true for all integers \(n \geqslant 1\)

Typical solution

Let \(n = 1\)

Then \(\left(\dfrac{1}{2}n(n + 1)\right)^2 = 1^2 = 1\) and \(\displaystyle\sum_{r=1}^{n} r^3 = 1\)

So the result is true for \(n = 1\)

Assume the result is true for \(n = k\)

Then

and

The result is true for \(n = 1\); if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction

for all integers \(n \geqslant 1\)

Typical solution

Try \(n = 1\):

and

LHS = RHS \(\therefore\) the rule is true for \(n = 1\)

Assume the rule is true for \(n = k\):

\(\therefore\) the rule is also true for \(n = k + 1\)

As the rule is true for \(n = 1\), and if true for \(n = k\) then it is also true for \(n = k + 1\), then by induction the rule is true for all integers \(n \geqslant 1\)

Typical solution

Let \(n = 1\) then \(\mathbf{M}^1 = \begin{bmatrix} 3 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \mathbf{M}\) so the result is true for \(n = 1\)

Assume the result if true for \(n = k\):

Hence true for \(n = k + 1\)

It is true for \(n = 1\). If it is true for \(n = k\) then it is true for \(n = k + 1\). Hence true by induction for all integers \(n \geqslant 1\)

Typical solution

Let \(n = 1\); then the formula gives

so the result is true for \(n = 1\)

Assume the result is true for \(n = k\):

Then \(u_{k+1} = \dfrac{5}{6 - \left(\dfrac{5^k - 5}{5^k - 1}\right)}\)

and the result also holds for \(n = k + 1\)

The formula for \(u_n\) is true for \(n = 1\); if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction \(u_n = \dfrac{5^n - 5}{5^n - 1}\) for \(n \geqslant 1\)

Typical solution

Assume it is true for \(n = k\)

where \(m\) is an integer

\(\therefore\) it is also true for \(n = k + 1\)

It is true for \(n = 1\). If it is true for \(n = k\) then it is true for \(n = k + 1\). Therefore, by induction, \(7^n - 3^n\) is divisible by 4 for all integers, \(n \geqslant 1\).

Typical solution

Try \(n = 1\):

\(\therefore\) true for \(n = 1\)

Assume true for \(n = k\)

\(\therefore\) it is also true for \(n = k + 1\)

True for \(n = 1\), and true for \(n = k \Rightarrow\) true for \(n = k + 1\)

Then, by induction, it is true for all integers \(n \geqslant 1\)

Typical solution

Invariant lines are \(y = 0\) and \(y = x + c\)

Typical solution

Invariant lines are \(y = 0\) and \(y = x\)

Typical solution

Typical solution

Let \(n = 1\) then \(\mathrm{f}(1) = 12 = 2 \times 6\)

so the result is true for \(n = 1\)

Assume the result is true for \(n = k\):

Then \(\mathrm{f}(k) = 6m\) for some integer \(m\)

\(3k^2 + 9k + 12\) is a multiple of 3

\(3k^2 + 9k + 12 = 3(k^2 + 3k) + 12\)

\(k^2 + 3k = k(k + 3)\) and one of \(k\) or \(k + 3\) is even

\(\therefore k^2 + 3k\) is even and \(3(k^2 + 3k + 4)\) is an even multiple of 3 and hence divisible by 6

\(\therefore \mathrm{f}(k + 1)\) is divisible by 6 if \(\mathrm{f}(k)\) is divisible by 6

We also know that \(\mathrm{f}(1)\) is divisible by 6, so by induction this completes the proof.

Notes

(corrected from the printed mark scheme: the printed line reads \(\mathrm{f}(k + 1) = k^3 + 6k^2 + 17k + 12 - (k^3 + 3k^2 + 8k)\); the \(+\,\mathrm{f}(k)\) term, which gives the \(6m\), was missing.)

Typical solution

\(\therefore\) it is true for \(n = 1\)

Assume it is true for \(n = k\)

\(\therefore\) it is also true for \(n = k + 1\)

True for \(n = 1\), and true for \(n = k \Rightarrow\) true for \(n = k + 1\), then by induction it is true for all integers \(n \geqslant 1\)

AG

Typical solution

AG

Notes

Throughout this question condone single line slips to sin or cos provided recovered.

B1: May see \(n = 1\) in addition but this mark is for \(n = 0\) only. Must see LHS = RHS \(\Rightarrow\) true for \(n = 0\) oe (but may appear later e.g. in conclusion)

M1: Sets up and uses inductive hypothesis by showing an intent to differentiate twice and using the inductive assumption at least once. Complete statement might not be seen until after first or even second differentiation.
Must have statement in terms of some other variable than \(n\) in all three places consistently.
Condone minor slips in notation e.g. \(\frac{\mathrm{d}^2\boldsymbol{f}}{\mathrm{d}x^2}\left(\frac{\mathrm{d}^{2k}\mathrm{f}}{\mathrm{d}x^{2k}}\right)\) for all but the final A mark.

M1: Differentiates using product rule. Condone sign errors, missing 2s, \(\frac{1}{2}\) instead of 2 but not \(\frac{1}{2}x^2\) as the derivative of \(x\).
May be seen separately. Could be done with exponential form but M1 awarded on correct return to hyperbolic form (which could be after later differentiation).

M1: Differentiates again. Same guidance. Or \(= 4^k(4k\cosh 2x + 4\cosh 2x + 4x\sinh 2x)\)
Allow both differentiations in one step if fully correct.

A1: Rewrites to the correct form.

A1: Clear conclusion for induction process with implication stated. Dependent on all previous M and A marks www and fully correct notation throughout.
Allow this mark if the basis case was given as \(n = 1\) only but their conclusion must match their basis case (i.e. \(n = 1 \Rightarrow n \geq 1\)).

Notes

M1: \(= 1024\) Chooses correct value of \(n\) and uses formula to find value for appropriate derivative when \(x = 0\).

M1: Using the correct rule for finding the coefficient of \(x^8\). This is independent of the previous mark and \(\mathrm{f}^{(8)}(0)\) need not be correct or even evaluated but must be \(x = 0\).

A1: Must be simplified. Do not accept eg 0.0254 etc but ISW once 8/315 seen unless it is clear that a different value is intended (eg 8 from multiplying by 315).
Allow with \(x^8\) e.g. \(\frac{8}{315}x^8\). Answer only with no working scores 0/3.

Notes

B1: Uses exponential definition of \(\sinh x\) with \(2x\) as argument…

M1: … and given general term (or complete series) for \(\mathrm{e}^x\) to try to find (coefficient of) \(x^7\) term of \(\sinh 2x\) (or \(x^8\) term of \(x\sinh 2x\)).

A1: No conclusion necessary. Must be in same form as answer to (b) (ie explicit verification) (but allow this mark for a correct answer if no solution in (b)).
Allow SC A1 if (b) and (c) both given as eg 0.0254. Allow with \(x^8\) e.g. \(\frac{8}{315}x^8\)

Notes

B1: BC.
The conclusion for LHS>RHS might appear in final statement.
Ignore calculations and statements relating to \(n = 51\).
Allow \(8 \times 10^{67} \gt 4.5 \times 10^{67}\)

M1: Setting up inductive hypothesis properly.

M1: Correctly using the inductive hypothesis
Could see:
\(k! \gt 20^k\)
\((k + 1)k! \gt (k + 1)20^k\)
Allow \(k\) to “slip” in \(20^k\) as long as intention is clear

A1: Must be sufficient justification
Withhold if \(k\) slipped

A1: Clear conclusion for inductive process. Must be 52.
A formal proof is required for full marks but other complete and correct proofs can get full marks.

Notes

B1: Base case shown to be true – condone when \(n = 0\), 11 – 1 – 1 = 9 or just 9, must say that 9 is divisible by 3 or show explicitly e.g. \(9 = 3 \times 3\).
Allow base case with \(n = 1\), \(11 \times 7 - 13 - 1 = 63\) or just 63, must say that 63 is divisible by 3 or show explicitly e.g. \(63 = 3 \times 21\).

M1*: Forms correct inductive hypothesis as an equation in terms of two different letters (oe in words e.g. \(11 \times 7^k - 13^k - 1\) is divisible by 3) – condone using \(n\) e.g. \(11 \times 7^n - 13^n - 1 = 3m\) for this mark.

M1dep*: Considers \(11 \times 7^{k+1} - 13^{k+1} - 1\) and use inductive hypothesis that \(11 \times 7^k - 13^k - 1 = 3m\) - must not be using \(n\) for \(k\).

A1: Shows using the inductive hypothesis that \(11 \times 7^{k+1} - 13^{k+1} - 1\) is a multiple of 3. Some common answers are \(3(-22 \times 7^k + 4 + 13m)\) or \(3(7m + 2 - 2 \times 13^k)\) or \(3(m + 22 \times 7^k - 4 \times 13^k)\).

A1: Conclusion, dependent on B1M1M1A1 and no errors seen in their proof. Do not award this mark if \(n = 0\) not used as base case. However, this mark can be awarded if \(n = 1\) used as base case and \(n = 0\) considered separately. Must mention, ‘\(n = k\) or P(\(k\))’, ‘\(n = k + 1\) or P(\(k + 1\))’, ‘\(n = 0\) or base case or P(0)’ (provided P(\(n\)) is defined) and ‘\(n \geqslant 0\) or all \(n\) oe’ condone ‘all integers’ but not ‘positive integers’ or similar incorrect statement.

Alternative method

B1: Base case shown to be true – condone when \(n = 0\), 11 – 1 – 1 = 9 or just 9, must say that 9 is divisible by 3 or show explicitly e.g. \(9 = 3 \times 3\).
Allow base case with \(n = 1\), \(11 \times 7 - 13 - 1 = 63\) or just 63, must say that 63 is divisible by 3 or show explicitly e.g. \(63 = 3 \times 21\).

M1*: Forms correct inductive hypothesis - condone using \(n\) e.g. \(\mathrm{f}(n) = 11 \times 7^n - 13^n - 1\) for this mark. Do not need to use \(\mathrm{f}(k)\) – can just state that \(11 \times 7^k - 13^k - 1\) is divisible by 3.

M1dep*: Considers correct expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) and simplifies to an expression of the form \(\pm a \times 7^k \pm b \times 13^k\) - must not be using \(n\) for \(k\).

A1: Shows using the inductive hypothesis that \(\mathrm{f}(k + 1)\) is divisible by 3 so must be considering an expression for \(\mathrm{f}(k + 1)\) and not just an expression for \(\mathrm{f}(k + 1) - \mathrm{f}(k)\).

A1: Conclusion, dependent on B1M1M1A1 and no errors seen in their proof. Do not award this mark if \(n = 0\) not used as base case. However, this mark can be awarded if \(n = 1\) used as base case and \(n = 0\) considered separately. Must mention, ‘\(n = k\) or P(\(k\))’, ‘\(n = k + 1\) or P(\(k + 1\))’, ‘\(n = 0\) or base case or P(0)’ (provided P(\(n\)) is defined) and ‘\(n \geqslant 0\) or all \(n\) oe’ condone ‘all integers’ but not ‘positive integers’ or similar incorrect statement.

Notes

B1: \(\mathbf{A}^1\) and \(a \times 1\) must both be seen explicitly.
Accept “\(= \mathbf{A}\)” instead of “therefore true when \(n = 1\)”

M1: Setting up the inductive hypothesis properly.

M1: Considering \(\mathbf{A}^{k+1}\) and using their inductive hypothesis. Could also consider \(\mathbf{AA}^k\)
Jumping to \(\mathbf{A}^{k+1} = \begin{pmatrix} 1 & a(k + 1) \\ 0 & 1 \end{pmatrix}\) without seeing the two matrices multiplied together first scores M0.

A1: Must see \(a + ak\) appear before being factorised.

A1: Clear conclusion for induction process. Must mention both basis case and that statement true for \(k\) implies true for \(k + 1\). BOD missing the word integer
If B0 not awarded as \(\mathbf{A}^1\) or \(a \times 1\) (or both) not seen allow A1 here (but must have attempt at base case)

Notes

M1: Differentiating correctly first derivative, or \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{a}{1 + at} = \left(a^{-1} + t\right)^{-1}\) etc

A1: (1st) For all; constants need not be evaluated and may be unsimplified, e.g. could see \((-1)^2\) or \((-1)^3\)

A1: (2nd) Could be \((-1)^{n+1}\) oe, or \((-1)^{n-1}(n - 1)!\left(a^{-1} + t\right)^{-n}\). Omission of factorial term can score a possible B1M1 in part (b)

Notes

*B1ft: Convincingly showing that their conjecture works for \(n = 1\) for their 1^st derivative in part (a); accept omission of working shown in 1^st line here

M1: Forming the inductive hypothesis and making it clear that the \((k + 1)\)^th derivative is the derivative of the \(k\)^th derivative

*A1: Differentiating and rewriting into correct form. Some intermediate working and/or justification must be seen. Could see substitution \(u = 1 + at\) etc

dep*A1: full correct argument

Notes

M1: Considering the seventh derivative.

A1: 5.625. Ignore attempt at units.

Notes

B1: Basis case. Must explicitly see 70 and state, or show, divisibility.
If \(n = 1\) (leading to \(98 = 7 \times 14\)) used and not corrected then allow this mark but withhold final mark.

M1: Statement of inductive hypothesis. Allow “\(= 14p\)” without further qualification

M1: Uses law of indices correctly to obtain expression in terms of \(8^k\) and no other exponential term
Could consider \(\mathrm{f}(k + 1) - \mathrm{f}(k)\) or similar

M1: Uses inductive hypothesis properly. Do not allow if eg \(32^k\) (ie law of indices must have been correctly used)
Allow for substitution of their \(n = k\) case into their \(n = k + 1\) case

A1: Simplification with sufficient working to establish and state divisibility for \(k + 1\).
Must either show \(14\times\) explicitly in each term or 14 is a factor

A1: Clear conclusion for induction process. See note for basis case above. Do not allow “true for all positive integers”.
Need to see if true for \(n = k\), then true for \(n = k + 1\) i.e. \(k\) and \(k + 1\)th case linked to \(n\).
Could see Proposition notation \((P_k \Rightarrow P_{k+1})\)
A formal proof by induction, with no gaps in logic, is required for full marks.
Full marks can be gained by using induction to prove that \(2 \times 8^n + 33\) is divisible by 7 for all \(n \geqslant 0\) and observing that \(4 \times 8^n + 66 = 2(2 \times 8^n + 33)\) is divisible by 2 and \(2 \times 8^n + 33\) and hence by both 2 and 7 and hence by 14

Notes

B1: Evidence of correct differentiation using product rule and a substitution \(n = 1\) into RHS

M1dep: Differentiate using the product rule

M1dep: Differentiate second time using the product rule

A1*: Must be equated to correct derivative form

A1dep: Dependent on previous A1 and the B1

Notes

B1: Basis case. Comparison must be explicit and correct
An assertion without calculation such as e.g. \(3^4 \gt 10 \times 4\) is insufficient for B1.
BOD statements such as “therefore true when \(n = 1\)”

M1: (1st) Inductive hypothesis set up

M1: (2nd) Considering for \(k + 1\) and using inductive hypothesis correctly
Asserting \(3^{k+1} \gt 10(k + 1)\) without justification gets M0.
Could compare \(3k\) and \(k + 1\)

A1: (1st) Showing enough working to establish statement for \(k + 1\). Must be justified but justification could be e.g. \(k \gt 1\).

A1: (2nd) Clear and complete conclusion, following a correct and complete proof with no incorrect statements. Must be \(n \geqslant 4\) not eg 0 or 1 for A1.
This mark must only be awarded if the language and notation in the whole proof and conclusion is correct.

Notes

B1: (first) BC

B1: (second) Allow this mark for any conjecture which works for \(n = 1\), 2, 3 and 4.

Notes

B1: Allow this mark even if the conjecture is wrong, provided that it works for \(n = 1\)

M1: (first) Must have statement in terms of some other variable than \(n\). Conjecture need not be correct.

M1: (second) Uses inductive hypothesis properly & expands

A1: AG. Manipulating terms correctly and convincingly to obtain required form. Some intermediate working must be seen and a clear conclusion must be given for the induction process.
A formal proof by induction is required for full marks.

Notes

B1: At least one intermediate step must be shown

M1: (1st) Must have statement in terms of some other variable than \(n\)

M1: (2nd) Uses laws of indices and then inductive hypothesis properly to eliminate either \(2^{3k}\) or \(3^k\) (not both)
or \(8 \times 2^{3k} - 3 \times (2^{3k} - 5p)\)

A1: (1st) AG. Further simplification to establish truth for \(k + 1\)
\(5(3p + 2^{3k})\)

A1: (2nd) Clear conclusion for induction process, following a correct proof by induction.
A formal proof by induction is required for full marks.

Notes

B1: Basis case – either \(4 \times 9\) or “36 is divisible by 9” given explicitly.

M1: For getting started (Sight of \(\mathrm{f}(r)\) is not necessary)

M1: For finding \(\mathrm{f}(r + 1)\)

A1: Each term as a multiple of 9

A1: Conclusion dependant on all other marks being earned

Notes

B1: Basis case. Comparison must be explicit and correct
Bod sight of “true for \(n=1\)”

M1: (1st) Inductive hypothesis set up
Might see \(2^{2k} = 4^k\) throughout

M1: (2nd) Considering for \(k + 1\) and using inductive hypothesis correctly
Allow M1 for use of inductive \(n = k\) step when showing \(P_{k+1} \to P_k\)

A1: (1st) Showing enough working to establish statement for \(k + 1\)
Might see \(\geqslant 10\) or \(\geqslant 9\) for \(\gt 9\).
A0 here if \(k \gt 9\) stated earlier
Do not allow if implication shown in the direction \(P_{k+1} \to P_k\)

A1: (2nd) Clear and complete conclusion

Notes

DR: Detailed reasoning required for this question

B1: Full details must be shown.

M1: (1st) Must have statement in terms of some other variable than \(n\).
Watch out for \(2^{k+1}\) appearing prematurely as bottom right element

M1: (2nd) Uses inductive hypothesis properly
or
\(\mathbf{M}^k\mathbf{M} = \begin{pmatrix} 1 & 3(2^{k+1} - 2) \\ 0 & 2^k \end{pmatrix}\begin{pmatrix} 1 & 6 \\ 0 & 2 \end{pmatrix}\)

M1: (3rd) Genuine attempt at matrix multiplication (ie columns multiplied into rows)
\(\begin{pmatrix} 1 & 6 + 2 \times 3(2^{k+1} - 2) \\ 0 & 2^k \times 2 \end{pmatrix}\)

A1: Simplification with sufficient working to establish truth for \(k + 1\)
Must see at least one stage of working between expression for \(\mathbf{M}^k\mathbf{M}\) (or \(\mathbf{MM}^k\)) and required expression for \(\mathbf{M}^{k+1}\) (e.g.:
\(\begin{pmatrix} 1 & 3(2 + 2 \times 2^{k+1} - 4) \\ 0 & 2^k \times 2 \end{pmatrix}\)
\(= \begin{pmatrix} 1 & 3(2^{(k+2)} - 2) \\ 0 & 2^{(k+1)} \end{pmatrix}\)

E1: Clear conclusion for induction process. Needs a fully correct proof which usually means all other marks awarded. Cannot be awarded if there are mistakes in the proof.
A formal proof by induction is required for full marks.
SC If B1M1M1M1 gained, but A0 because of lack of intermediate step of working then allow SC B1 for fully correct ending statement

Notes

M1: (1st) Basis case. Must explicitly state divisibility (\(49 \div 7 = 7\) is OK).
Do not condone \(k = 0\) unless later stated \(1 \gt 0\)

M1: (2nd) Statement of inductive hypothesis. Allow “\(= 7p\)” without further qualification.

M1: (3rd) Might not all be done before next step. Do not allow if e.g. \(45^k\).
Needs to have a \(2^{k+1}\) or \(5 \times 9^k\) so that the \(n = k\) case can be used.

M1: (4th) Uses inductive hypothesis properly. Do not allow if e.g. \(45^k\).
Do not allow M1 if both replacements made unless recovered later

A1: Simplification with sufficient working to establish divisibility for \(k + 1\)

E1: Clear conclusion for induction process.
A formal proof by induction is required for full marks.

Notes

B1: no more simplified in initial step than \(\frac{(-1)^2(0!)}{1 + x}\); condone missing brackets

M1: differentiating expression for \(\mathrm{f}^{(k)}(x)\); condone bracketing errors or a sign slip only

A1: condone \(\frac{(-1)^{(k+2)}k!}{(1 + x)^{k+1}}\). Do not condone bracketing errors. M1 step must be seen.

A1: www. \(\frac{(-1)^{(k+1)+1}((k + 1) - 1)!}{(1 + x)^{k+1}}\) must have been seen to score this mark. \(n = 1\) must have been considered. Cannot score A0A1.

Notes

M1: using result from (a) to find at least two derivatives (can include general term). May be embedded in coefficients. Must include powers of \(-1\) and factorials in each term; condone missing division by 1. Allow one slip.

M1: considering coefficient of general term and simplifying using \(\frac{(n - 1)!}{n!}\)

A1: AG complete argument, www. Must clearly be a sum of infinitely many terms. Must have considered \(\mathrm{f}(0) = 0\).

Alternative method

M1: use of Maclaurin series with result from (a) used for all terms other than \(\mathrm{f}(0)\)

A1: \(\frac{(n - 1)!}{n!} = \frac{1}{n}\) seen or used

A1: AG

Notes

M1: Do not accept “sheaf”

A1: Accept \(x\)-axis is a line of invariant points (not an invariant line only). “Shear parallel to \(x\)-axis” is insufficient.
Accept alternative mappings, e.g. \((1, 1)\) to \((1 + \lambda, 1)\).
Do not accept shear factor.

Notes

(b)(ii)

B1: FT their determinant. Condone area scale factor = 1.

B1: FT their determinant. Do not accept “orientation not reversed”.

Notes

B1: \(1 \times \lambda\) must be seen. “True when \(n = 1\)” could appear later.

M1: \(\mathbf{M}^{k+1} = \mathbf{M}^k\mathbf{M}\) or \(\mathbf{M}^{k+1} = \mathbf{M}\mathbf{M}^k\) could be used.

A1*: Required matrix with intermediate step seen

A1dep: \(n = 1\) must have been considered

Notes

B1: Accept \(x\)-axis is a line of invariant points (not an invariant line only).
Accept alternative mappings, e.g. \((1, 1)\) to \((1 + n\lambda, 1)\).
Do not accept shear factor

Notes

B1: check true for \(n = 1\)

A1: (final) cao dep previous A1

Notes

B1: [ \(= \mathbf{I}\) may be inferred from ‘student is correct’]

B1: (3rd) Must show evidence for inverse
e.g. calculate det or show \(\mathbf{MM}^{-1} = \mathbf{I}\)

Notes

M1: (1st) or \(3^k = 8^k - 5m\) or \(8^k - 3^k = 5m\) or \(8^k - 3^k\) div 5

M1: (2nd) assumption used

A1: (1st) successful completion
\(3^k = 8^k - 5m\) used \(\Rightarrow 5(8^k + 3m)\) div 5 www
\(8^k - 3^k = 5m\) used \(\Rightarrow 5(8m + 3^k)\) div 5 www

A1: (2nd) must receive all previous marks for this to be awarded

Alternative method

M1: (2nd) considering difference between \(u_{k+1}\) and \(u_k\)

A1: (2nd) must receive all previous marks for this to be awarded

Notes

B2: Allow B1 if \(\mathbf{M}^2\) correct

Notes

B1: oe – allow correct unsimplified expressions
allow if correct expression is seen in part (c)

Notes

M1: or \(\mathbf{M} \times \mathbf{M}^k\)

A1: (2nd) or using target expression

B1: (2nd) dep first three marks gained. Must have if ... then... (oe)

Notes

M1: Intermediate step seen

B1cao: Must receive all 4 previous marks for this to be awarded

Notes

M1: (2nd) Factoring \(n\) or \(n + 1\)

A1: (2nd) NB AG must show previous step

Notes

B1: checking \(n = 1\)

M1: (1st) condone \(\displaystyle\sum_{r=1}^{k+1} k(k + 2) = \ldots\)

M1: (2nd) factorising

A1: (2nd) or equals target expression if given

A1cao: dep A1* must have both statements

Notes

A1: Dependent on all previous marks awarded

Notes

B1: (2nd) soi from correct next step
but \(\displaystyle\sum_{r=1}^{k} k \times 2^{k-1}\) is B0

M1: condone missing brackets if intention is clear subsequently

B1dep: dep A1*

Notes

A1*: Or target seen

B1: Final B mark only awarded if all previous marks awarded

Notes

B1*dep: dep A1*

Notes

M1: (Assume) or \(5^k + 2 \times 11^k = 3m\)

M1: substituting for \(5^k\)

A1: or \(15m + 12 \times 11^k\)

A1*, A1dep: dep * marks

Alternative for the 4th and 5th marks

First: substituting for \(11^k\), or \(33m - 6 \times 5^k\); \(5.5^k + 11(3m - 5^k)\)

Second: adding \(u_k\) to \(u_{k+1}\)

Notes

M1: (1st) condone notation errors

M1: (2nd) adding \(\dfrac{1}{3^{k+1}}\) to \(\dfrac{1}{2}\left(1 - \dfrac{1}{3^k}\right)\)

A1: combining fractions

E1dep: dep A1*

Notes

M1: (2nd) or \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 2^k - 1 \\ 0 & 2^k \end{pmatrix}\)

A1*: or \(\begin{pmatrix} 1 & 2^k - 1 + 2^k \\ 0 & 2^{k+1} \end{pmatrix}\)

M1dep*: must clearly ‘assume’ \(n = k\)
‘true for \(n = k\) and \(k + 1\)’ is M0

A1dep*: must have established truth for \(n = 1\) for final mark
but need not re-state it at the end