Notes

M1: Starts the method by writing the correct sum and splitting into two sums, condone use of \(n\) instead of \(r\)

M1: Substitutes in at least one standard formula

A1: Fully correct unsimplified expression

dM1: Dependent on the previous method mark. Factorises out at least \(n\).

A1*: Achieves the printed answer with no omission or errors, factorisation of (\(n\) + 1) term needed cso

Notes

M1: Sets the printed answer equal to \(22n\), forms a quadratic/cubic and solves to find a non zero value for \(n\). If an incorrect equation please check that their value for \(n\) is correct for their equation if no method shown.

A1: Selects the correct positive value of \(n\). If \(n = -13\) appears this must be rejected

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

M1: A complete strategy to write as partial fractions

dM1: Dependent on previous method mark. Completes the method of differences process, must have a minimum substitution of \(r = 2,\ r = 3,\ r = n - 1\) and \(r = n\). Condone the use of \(r\) for \(n\) for this mark.
If pairs of terms are not seen in each substitution for \(r\), this mark can be implied by sight of \(2 + 1 - \dfrac{2}{n} - \dfrac{2}{n + 1}\) following correct partial fractions.

A1: Correct ‘fractions’ from the beginning and end that do not cancel. Accept exact equivalents. Must be in terms of \(n\)

M1: Combines all their ‘fractions’ (at least two algebraic fractions) over a correct common denominator, leading to a quadratic in the numerator, where brackets may not be expanded. Condone the use of \(r\) for \(n\) for this mark.

A1: Since the form of the solution is given, only accept \(\dfrac{(3n + 2)(n - 1)}{n(n + 1)}\) and not equivalents, but isw
Do not accept values for \(p\) and \(q\) written alone.

Note: If they start with \(r = 0,\ r = 1\) or \(r = 3\), the maximum they can score is M1M0A0M1A0
Similarly, if they use \(r = n + 1\) the maximum they can score is M1M0A0M1A0

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

M1: Recognises the need to find partial fractions and applies a correct method leading to finding values for \(A\) and \(B\)
Allow a slip when finding the constants

A1: Correct partial fractions seen at any stage. Not just values for \(A\) and \(B\) listed

Note: Proof by induction will not score the next 4 marks.

M1: Starts the process of finding terms at the start and at the end, in order to establish the non-cancelling terms
Must have attempted a minimum of \(r = 1,\ r = 2,\ \ldots\ r = n - 1\) and \(r = n\), this may be implied by their correct non-cancelling terms.
Follow through on their values of \(A\) and \(B\). Look for

\(r = 1 \rightarrow \dfrac{A}{5} - \dfrac{B}{7}\)     \(r = 2 \rightarrow \dfrac{A}{6} - \dfrac{B}{8}\)
\(r = n - 1 \rightarrow \dfrac{A}{n + 3} - \dfrac{B}{n + 5}\)     \(r = n \rightarrow \dfrac{A}{n + 4} - \dfrac{B}{n + 6}\)

A1: Correct non-cancelling terms which may be listed separately.
Correct fractions from the beginning and end that do not cancel stated.

M1: Combines ‘their’ fractions of the form \(p + \dfrac{q}{n + 5} + \dfrac{r}{n + 6}\) over a correct common denominator which does not need to be the lowest common denominator and obtains a quadratic expression in the numerator.

A1: Correct answer.

Note: If they start with \(r = 0\) the maximum they can score is M1A1M0A0M1A0

Notes

M1: Substitutes at least one of the standard formulae into their expanded expression

A1: Fully correct expression

dM1: Attempts to factorise \(\dfrac{1}{12}n(n + 1)\) or \(\dfrac{1}{3}n(n + 1)\) having used at least one standard formula correctly at any stage. Dependent on the first M mark.
If they show no method for factorising (use a calculator) they can go from \(3n^3 + 10n^2 + 9n + 2 = (n + 1)(n + 2)(3n + 1)\)

A1: Obtains the correct expression or the correct values of \(a\) and \(b\), with no errors seen

Notes

M1: Uses the result from part (a) and adopts a correct strategy by attempting \(\displaystyle\sum_{r=1}^{3k} r^2(r + 1) - \sum_{r=1}^{k} r^2(r + 1)\)

M1: Factorises out at least \(k(3k + 1)\) at any stage, could be done by inspection

A1: Obtains the correct expression with no errors seen.

Note If a candidate does not use part (a) but restarts they can still score marks for the same reasons. If unsure please send to review

Notes

M1: Uses the given equation, substitutes their answer from part (b) and simplifies to either reach

• \(Ak^4 + Bk^3 + Ck^2 + Dk\{= 0\}\)
• \(k\left(Ak^3 + Bk^2 + Ck + D\right)\{= 0\}\) or \((3k + 1)\left(Ak^3 + Bk^2 + Ck\right)\{= 0\}\)
• \(Ak^3 + Bk^2 + Ck\{= 0\}\)
• a 3TQ or \(k(3k + 1)\left(Ak^2 + Bk + C\right)\{= 0\}\)

this can be implied by a correct value for \(k\) if left unsimplified

M1: Solves their equation as long as solving their \((b) = 192k^3(3k + 1)\) to find a non zero value for \(k\), including by calculator. You may need to check this.

A1: Selects the appropriate correct answer of \(k = 5\). Any other solutions must be clearly rejected.
Note: Correct answer with no working is M0M0A0

Notes

B1: Correct expanded expression.
M1: Substitutes at least one of the standard formulae into their expanded expression.
A1: Fully correct unsimplified expression.

dM1: Dependent on previous method. Attempts to factorises out \(n\). Must have a \(n\) in every term. Condone a slip with one term as long as the intention is clear.

A1: Achieves the correct answer, with a correct intermediate line of working. cso

Note If uses \(\sum 1 = 1\) scores B1 M1 A0 M0 A0

An attempt at proof by induction may score B1 only

Notes

B1: Correct summation formula for the sum of the squares of all positive odd three-digit integers including limits. This can be implied by later work.

M1: Uses the answer to part (a) and \(\displaystyle\sum_{r=p}^{q} (2r - 1)^2 = \sum_{r=1}^{q} (2r - 1)^2 - \sum_{r=1}^{p-1} (2r - 1)^2\) where \(p\), \(q\) are numerical and \(q > p\), to find a value. There must be some indication of the sum that they are finding or the correct values for \(p\) and \(q\).
States \(\displaystyle\sum_{r=1}^{500} (2r - 1)^2 - \sum_{r=1}^{50} (2r - 1)^2\) implies B1

States \(\dfrac{500}{3}\left(4(500)^2 - 1\right) - \dfrac{50}{3}\left(4(50)^2 - 1\right)\) this scores B1 (implied) and M1

A1: Correct value

Note \(\displaystyle\sum_{r=51}^{500} (2r - 1)^2 = 166499850\) or correct answer only scores B1 M0 A0, must be evidence of using the answer to (a)

Notes

M1: This mark is for stating that all the odd terms are negative (or subtracted) and all the even terms are positive (or added) or for showing this by writing down at least 4 terms as above. May be achieved via rhs e.g. \(\displaystyle\sum_{r=1}^{n} \left(\mathrm{f}(2r) - \mathrm{f}(2r - 1)\right) = \mathrm{f}(2) - \mathrm{f}(1) + \mathrm{f}(4) - \mathrm{f}(3) + \ldots\)

A1*: A full and convincing argument that shows the equivalence of both sides of the equation. E.g. clearly demonstrates how the terms separate as \(\mathrm{f}(2) + \mathrm{f}(4) + \ldots + \mathrm{f}(2n)\) for the even powers and \(\mathrm{f}(1) + \mathrm{f}(3) + \ldots + \mathrm{f}(2n - 1)\) for the odd powers and concludes \(\displaystyle\sum_{r=1}^{n} \left(\mathrm{f}(2r) - \mathrm{f}(2r - 1)\right)\,*\)
Need to see:

• the even terms grouped as e.g. \(\mathrm{f}(2) + \mathrm{f}(4) + \ldots + \mathrm{f}(2n)\)
• the odd terms grouped as e.g. \(\mathrm{f}(1) + \mathrm{f}(3) + \ldots + \mathrm{f}(2n - 1)\)
• a conclusion

See below for some examples.

Notes

M1: Expands summand fully and applies \((-1)^{2r} = 1\) or \(r(-1)^{2r} = r\) at some point in their solution.
Need not split the summation for this mark, \(\sum r\left(1 + k(-1)^r r + mr^2\right)\) can be accepted where \(k\) and \(m\) are 2 or 4.)

M1: Applies the correct formulae for sum of integers and sum of cubes to the \(r\) and \(mr^3\) terms with \(2n\)

M1: Applies result of part (a) to the \((-1)^r\) term. Condone minor slips but the structure should be correct e.g.
\(\lambda\displaystyle\sum_{r=1}^{2n} (-1)^r r^2 = \sum_{r=1}^{n} \left(\alpha(2r)^2 - \beta(2r - 1)^2\right)\) but not e.g. \(\lambda\displaystyle\sum_{r=1}^{2n} (-1)^r r^2 = \sum_{r=1}^{n} \left(\alpha(2r)^2 - \beta\left((2r)^2 - 1\right)\right)\), \(\alpha, \beta \gt 0\)

B1: Correct expression for \(4\displaystyle\sum_{r=1}^{2n} (-1)^r r^2\) which may be unsimplified e.g. \(4\left(4\dfrac{n}{2}(n + 1) - n\right)\) as shown or e.g. \(8n^2 + 4n\), \(\dfrac{16}{2}n(n + 1) - 4n\), etc. Depends on previous method mark.

dM1: Depends on all previous M marks. Takes out a factor of \(n(2n + 1)\) from a quartic expression.
To score this mark, there must be a factor of \(n\) and an obvious factor of \((2n + 1)\) if they leave the expressions inside the outer brackets factorised. However, this mark can be scored if they expand fully, take out a factor of \(n\) and then attempt to e.g. divide by \((2n + 1)\) or e.g. use inspection to determine the quadratic factor. Must reach at least \(= n(2n + 1)\left(An^2 + \ldots n\right)\) where \(A\) is the coefficient of their cubic expression.

A1*: Correct answer obtained with no errors seen and suitable intermediate steps shown.
Note that it is acceptable to go from \(n\left(16n^3 + 16n^2 + 14n + 5\right)\) to \(n(2n + 1)\left(8n^2 + 4n + 5\right)\)

Notes

M1: Attempts to split the sum to a difference upper limit 50 and lower limit 13. May be implied.
Condone e.g. \(\displaystyle\sum_{r=14}^{50} r\left((-1)^r + 2r\right)^2 = \sum_{r=1}^{50} \mathrm{f}(r) - \sum_{r=1}^{13} \mathrm{f}(r)\) or even \(\displaystyle\sum_{r=14}^{50} r\left((-1)^r + 2r\right)^2 = \sum_{r=1}^{50} - \sum_{r=1}^{13}\)

M1: Splits expression with even upper limits for both sums and evaluates an appropriate balancing term. The balancing term must be correct for their method but ignore whether it is added or subtracted for this mark.

M1: Attempts to use the result from (b) at least once correctly with an integer – so using \(n = 25\) with upper limit 50, or \(n = 6\) with upper limit 12 e.g. \(25 \times 51 \times 5105\) or \(6 \times 13 \times 317\). Use of their upper limits in the formula is M0

A1: 6476024 from correct work. Question says “hence” so use of the result from (b) must be clear.

Notes

M1: Applies the formulas for sum of integers and sum of squares of integers to the summation.

A1: Correct unsimplified expression for the sum, including the \(25n\)

M1: Expands and factors out the \(n\) or ½n

A1: Correct proof, no errors seen.

Notes

M1: Writes out first few terms of the sum, at least 3, and identifies the repeating pattern, e.g. through bracketed terms or stating sum repeat every three terms oe.

A1: Correct explanation identifying \(-\sqrt{3}\) is the sum of each group of three terms, so with \(k\) lots of three terms the sum is \(-k\sqrt{3}\)

Notes

M1: Applies formula from (a) to left-hand side as a difference of two summations with either 4 or 5 as the limit on the second sum.

A1: Correct expression for the left-hand side in terms of \(n\)

B1: Correct expression for the sum on the right-hand side, allow if it arises from lower limit 6 used instead of 5 as the 6^th term is zero. May subtract the first few terms directly from the work in (b).

M1: Both sides expanded and terms gathered to reach a simplified cubic equation for \(n\) with no other unknowns (may not have factor of \(n\) if errors made, which is fine for the method mark).
This mark is not dependent on any previous marks and can be awarded as long as there is an attempt at both sides of the equation and an attempt at squaring their \(\displaystyle\sum_{r=6}^{3n} r\tan(60r)^\circ\).
If divides through by \(n\) this mark is awarded for a 3TQ

M1: Solves their cubic equation, which may be via calculator (so may need to check values). They may divide by \(n\) and solve a quadratic. Condone decimal roots truncated or rounded

A1: Selects the correct value of \(n\) to give 49 as the only non-trivial answer. The value 3 must be rejected as summation on left undefined for this value, but accept if 0 and 49 are given (since both sides evaluate to 0 for \(n = 0\) depending on one’s interpretation of summations).

Notes

M1: Uses the subtraction laws of logs to start the method of differences process.

dM1: Demonstrates the method of differences process, should have a minimum of e.g \(r = 2\), \(r = 3\), \(r = 4\), \(r = n - 1\) and \(r = n\) shown -- enough to establish at least one cancelling term and all non-disappearing terms though the latter may be implied by correct extraction if only the first few cases are shown. Allow this mark if an extra term for \(r = 1\) has been included.

A1: Correct terms that do not cancel - must not contradict their list of terms so e.g. if \(r = 1\) was included, then A0A0 follows. The \(\ln 1\) may be included for this mark.

A1*: Achieves the printed answer, with no errors or omissions and must have had a complete list (as per dM1) before extraction (but condone missing brackets on ln terms). If working with \(r\) throughout, they must replace by \(n\) to gain the last A, but all other marks are available.

NB For attempts at combining log terms instead of using differences, full marks may be awarded for the equivalent steps, but attempts that do not make progress in combining terms will score no marks.

Notes

Condone a bottom limit of 0 or 1 being used throughout part (b).

M1: Attempts to split into (the sum up to 100) – (the sum up to \(k\)) where \(k\) is 49, 50 or 51 and apply the result of (a) in some way. Condone slips with the power.

M1: Having attempted to apply (a), uses difference and power log laws correctly to reach an expression of the required form.

A1: Correct answer. Accept equivalents in required form, such as \(35\ln\dfrac{5050}{1275}\)

Notes

(a)
B1: See scheme

Notes

(b)(i)
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into at least 3 terms of the series and applies de Moivre’s theorem.
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into their answer to part (a) and rationalises the denominator.
M1: Equates the imaginary terms.
M1: Multiplies out the denominator and simplifies by using the identity \(\cos^2\theta + \sin^2\theta = 1\)
A1*: cso. Achieves the printed answer having substituted \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into 4 terms of the series.

Alternative
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into at least 3 terms of the series and applies de Moivre’s theorem.
M1: Substitutes \(z = \dfrac{1}{2}\mathrm{e}^{\mathrm{i}\theta}\) into their answer to part (a) and rationalises the denominator.
M1: Uses \(\mathrm{e}^{-\mathrm{i}\theta} = \cos\theta - \mathrm{i}\sin\theta\) and \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\) to express in terms of \(\sin\theta\) and \(\cos\theta\)
M1: Select the imaginary terms.
A1*: cso Achieves the printed answer having substituted \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into 4 terms of the series.

(corrected from the printed mark scheme: in the alternative, the denominator \(1 - \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta} - \frac{1}{2}\mathrm{e}^{-\mathrm{i}\theta} + \frac{1}{4}\) is printed with \(\frac{1}{4}\mathrm{e}^{\mathrm{i}\theta}\) and \(\frac{1}{4}\mathrm{e}^{-\mathrm{i}\theta}\))

Notes

(b)(ii)
M1: Setting the real part of the series \(= 0\) and rearranges to find \(\cos\theta = \ldots\)
A1: See scheme

Alternative 1
M1: Rearranges the real part so that \(\cos\theta\) only appears once
A1: Uses \(-1 \leqslant \cos\theta \leqslant 1\) to show that the sum must always be positive so must contain a real part

Alternative 2
M1: Sets sum as purely imaginary and rearranges to make \(z\) the subject
A1: Shows a contradiction and draws an appropriate conclusion

(corrected from the printed mark scheme: in Alternative 1 the expression \(\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta}\) is the real part but is printed as “Imaginary part”, and the note says “Rearranges imaginary part”; the bounds \(\dfrac{1}{6}\) and \(\dfrac{3}{2}\) are reached at \(\cos\theta = -1\) and \(\cos\theta = 1\), so they are printed as strict inequalities but should be \(\leqslant\))

Notes

(a)
M1: A complete attempt to find the sum of the cubes of \(n\) odd numbers using three of the standard summation formulae.
M1: Expands \(\sum_{r=1}^{n}(2r - 1)^3\) or \(\sum_{r=0}^{n-1}(2r + 1)^3\) and splits into four appropriate sums.
M1: Applies the result for at least three summations \(\sum_{r=0}^{n-1}r^3, \sum_{r=0}^{n-1}r^2, \sum_{r=0}^{n-1}r\) and \(\sum_{r=0}^{n-1}1\) or \(\sum_{r=1}^{n}r^3, \sum_{r=1}^{n}r^2, \sum_{r=1}^{n}r\) and \(\sum_{r=1}^{n}1\) as appropriate to their expansion provided that there is an attempt at cubing some values.
A1: Correct unsimplified expression.
A1*: Multiplies out to achieve a correct intermediate expression which clearly leads to the correct expression. cso

Special case: If uses \(\sum_{r=1}^{n}(2r + 1)^3\) leading to \(= 8\dfrac{n^2}{4}(n + 1)^2 + 12\dfrac{n}{6}(n + 1)(2n + 1) + 6\dfrac{n}{2}(n + 1) + n\) max score is M1 M0 M1 A1 A0

Notes

(b)
M1: Uses the answer to part (a) to find the sum of the cubes of the first \(N + 10\) odd numbers minus the sum of the first \(N\) odd numbers and sets equal to 99800 or equivalent.
A1: Correct simplified cubic equation.
dM1: Uses their calculator to solve their cubic equation, dependent on previous method mark.
A1: cao

Notes

B1: Correct volume of a block

M1: Expressing the total volume of all \(n\) blocks as a series in terms of \(r\), \(r^2\) and \(r^3\)

M1: Substitutes at least one of the standard formulae into their volume.

dM1: Attempts to factorise \(\dfrac{1}{4}n(n + 1)\) having used at least one standard formula correctly. Each term must contain a factor of \(n(n + 1)\)

A1*: Obtains the printed result with no errors seen, no bracketing errors and following from \(V = \dfrac{1}{4}n(n + 1)\left[n^2 + 5n + 6\right]\) o.e.

Note: Going from \(\dfrac{1}{4}n\left(n^3 + 6n^2 + 11n + 6\right)\) to \(\dfrac{1}{4}n(n + 1)(n + 2)(n + 3)\) with no reasoning shown scores dM0 A0

Notes

M1: Sets the printed answer \(= n^4 + 6n^3 - 11710\), simplifies, collects terms and uses their calculator to solve a quartic equation to find a value for \(n\).

A1: Selects \(n = 10\) or states that there are 10 blocks from a correct equation

Notes

M1: Selects the correct procedure for finding the mean (\(\bar{x}\)), attempting sum and dividing by \(n\).

M1: Splits the sum and applies the formulae for \(\sum r\) (accept \(7 + 3\dfrac{n}{2}(n + 1)\) here)
Or uses arithmetic series formula \(\dfrac{1}{2}n(a + l)\) with \(a = 10\) and \(l\) an attempt at \(7 + 3 \times n\), or \(\dfrac{n}{2}(2a + (n - 1)d)\) with \(a = 10\) and \(d = 3\).

A1*: Correct work proceeding to the answer with an intermediate step shown.
Special case: Award M0M1A0 for candidates who use \(\dfrac{1}{2}(a + l)\) or equivalent without justification of the division by \(n\).

Notes

(In the scheme, the single-underlined terms are for the M1 and the double-underlined term is for the B1.)

M1: Correct overall strategy to get as far as the variance of marbles in the collection. The attempt at variance should be recognisable (though allow e.g. sign slips in the formula for this mark) and an attempt, however poor, at \(\sum(7 + 3r)^2\) must have been made

B1: Correct value for the mean for 85 marbles (accept as a single fraction, \(\frac{272}{2}\)). If a student works algebraically until the last step, a correct final answer will imply this mark.

M1: Expands brackets and applies summation formulae for \(\displaystyle\sum_{r=1}^{n}r\) and \(\displaystyle\sum_{r=1}^{n}r^2\) to their expression, either in terms of \(n\) or with \(n = 85\) but must have correct limits. Allow for obtaining an expression of the correct form for Way 2 if the mean is kept in terms of “\(n\)”.
This mark is for correct application of these two summation formula on an attempt at \(\displaystyle\sum_{r=1}^{n}(7 + 3r)^2\) so accept even if this is not part of an attempt at the variance.

B1: Correct use of \(\displaystyle\sum_{r=1}^{n}1 = n\) in their expression (must be correct limits).

M1: Correctly applies variance or standard deviation formula with \(n = 85\), their attempt at \(\sum x^2\) (which need not be using \(7 + 3r\) or correct limits) and their mean. Accept use of the sample variance/standard deviation (dividing by \(n - 1\))
For reference the variance formula is\[\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = \left(\frac{1}{n}\sum_{i=1}^{n}x_i^2\right) - \bar{x}^2\]where \(x_r = 7 + 3r\) here, or accept for sample variance\[\sigma^2 = \frac{1}{n - 1}\sum_{i=1}^{n}(x_i - \bar{x})^2 = \left(\frac{1}{n - 1}\sum_{i=1}^{n}x_i^2\right) - \frac{n\bar{x}^2}{n - 1}\]

A1: Correct standard deviation to 1 decimal place. If sample standard deviation is used, the answer will be 74.0 g to 1 d.p. (74.04…)

Note: Question specifies use of summation formula and so these must be seen for the 2nd M and 2nd B mark. However, if just 2032690 appears from a calculator all other marks are available.

Notes

M1: A complete strategy to find \(A\), \(B\) and \(C\) e.g. partial fractions. Allow slip when finding the constant but must be the correct form of partial fractions and correct identity.

M1: Starts the process of differences to identify the relevant fractions at the start and end. Must have attempted a minimum of \(r = 0,\ r = 1,\ \ldots\ r = n - 1\) and \(r = n\)

Follow through on their values of \(A\), \(B\) and \(C\). Look for

\(r = 0 \rightarrow \dfrac{A}{1} - \dfrac{B}{2} + \dfrac{C}{3} \qquad r = 1 \rightarrow \dfrac{A}{2} - \dfrac{B}{3} + \dfrac{C}{4}\)

\(r = n - 1 \rightarrow \dfrac{A}{n} - \dfrac{B}{n + 1} + \dfrac{C}{n + 2} \qquad r = n \rightarrow \dfrac{A}{n + 1} - \dfrac{B}{n + 2} + \dfrac{C}{n + 3}\)

A1: Correct fractions from the beginning and end that do not cancel stated.

M1: Combines all ‘their’ fractions (at least two algebraic fractions) over their correct common denominator, does not need to be the lowest common denominator (allow a slip in the numerator).

A1: Correct answer.

Note: if they start with \(r = 1\) the maximum they can score is M1M0A0M1A0

Note: Proof by induction gains no marks

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

(a) Do not allow proof by induction (but the B1 could score for \((3r - 2)^2 = 9r^2 - 12r + 4\) if seen)

B1: Correct expansion

M1: Substitutes at least one of the standard formulae into their expanded expression

A1: Fully correct expression

dM1: Attempts to factorise \(\dfrac{1}{2}n\) having used at least one standard formula correctly. Dependent on the first M mark and dependent on there being an \(n\) in all terms.

A1*: Obtains the printed result with no errors seen

Notes

M1: Uses the result from part (a) by substituting \(n = 4\) and subtracts from the result in (a) in order to find the first sum in terms of \(n\).

M1: Identifies the periodic nature of the second sum by calculating terms. This may be implied by a sum of 14.

A1: Uses their sum and the given result to form the correct 3 term quadratic

M1: Solves their three term quadratic to obtain at least one value for \(n\)

A1: Obtains \(n = 29\) only or obtains \(n = 29\) and \(n = -\dfrac{88}{3}\) and rejects the \(-\dfrac{88}{3}\)

Typical solution

Typical solution

Typical solution

Notes

(corrected from the printed mark scheme: the M1 rows print “Substitutes \(r = 2c\) … \(r = c - 1\)”; the substitution is into the \(n\) of the part (b) result.)

Typical solution

Typical solution

Solutions to \(2n^2 + 2n - 10^5 = 0\) are 223.1 and \(-224.1\)

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

By de Moivre’s theorem,

Typical solution

Typical solution

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

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Notes

(corrected from the printed mark scheme: the first line of the typical solution prints \(\displaystyle\sum_{r=1}^{n}\left(1 + \frac{2}{r}\right)\), with the \(\ln\) missing.)

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

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

2450

Typical solution

Typical solution

Typical solution

Sixth powers: \(1 + 64 + 729 = 794\)

Required total:

Typical solution

Typical solution

Typical solution

Typical solution

since both denominators are positive.

The equation

has a positive and a negative root (since \(12 - 5k \lt 0\))

Since the root is positive, if

then

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

AG

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

M1: For correct identity oe e.g. \(P(2r + 1) + Q(2r - 1) \equiv 1\) following \(\dfrac{1}{(2r - 1)(2r + 1)} \equiv \dfrac{P}{2r - 1} + \dfrac{Q}{2r + 1}\) - condone if in terms of \(n\) – can be implied from correct partial fractions

A1: oe, allow for \(\dfrac{1}{(10r - 5)(10r + 5)} \equiv \dfrac{A}{10r - 5} + \dfrac{B}{10r + 5}\) followed by \(A = \dfrac{1}{10}, B = -\dfrac{1}{10}\) - condone if in terms of \(n\)

M1: Uses method of differences for their two term partial fractions of the correct form (where their \(A\) and \(B\) must have different signs). Must show at least the cases for \(r = 1, 2\) and \(n\) or \(r = 1, n - 1\) and \(n\) (so either first two terms and last term or first term and last two terms) with subtraction between terms
Condone missing or incorrect factor of \(\dfrac{1}{50}\)

B1: oe e.g. \(\dfrac{1}{50} - \dfrac{1}{100n + 50}\)

A1: www - must explaining why the sum is less than \(\dfrac{1}{50}\) - must consider the sign of \(\pm\dfrac{1}{2n + 1}\) correctly (oe) - dependent on all previous marks

Notes

B1*: soi

M1dep*: Using their \(S_k\) from part (a) which must be of the form \(a - \dfrac{b}{2n + 1}\) where \(a\) is non-zero and \(b \gt 0\), in the given equation, to get an equation in \(k\) e.g. \(\dfrac{24}{1225} = \dfrac{1}{50}\left(1 - \dfrac{1}{2k + 1}\right)\)

A1: cao - SCB1 for the correct answer with no working

Notes

B1: May be seen embedded in a sum

M1: Using the correct formula for sum of cubes with an upper limit (between 26-720 but not 216) substituted

A1: Any one of these seen correct
May see \(1 + 8 + 27 + 64 + 125 = 225\)

A1: Correct answer with no working is 0/4

Alternative method

B1: May be seen embedded in a sum (e.g. 75 as upper limit)

M1: Expressing the required sum as a single sum and multiplying out so that standard formulae can be applied (allow one slip in coefficients). Ignore limits for this mark.

A1: 3 out of 4 terms correct (must include \(125 \times 75\))

A1: Correct answer with no working is 0/4

Notes

M1: DR. Correctly setting up \(S\) as a sum of odd squares (condone incorrect limit combination: allow 0 or 1 for lower and 389 or 390 or 399 or 400 for upper). Treat 399 or 400 as MR (799 for 779).

M1: Expanding and separating (with their limits). Term “1” must be dealt with correctly although condone poor notation if their 390 appropriately seen later). M0M1 is possible

B1FT: FT only 389 or 390. Correctly using the formula for the sum of squares up to and including 390 (or 389)
or 19 697 015 from 0 or 1 up to 389

B1FT: FT only 389 or 390. Correctly using the formula for the sum of integers up to and including 390 (or 389) and “sigma 1 = \(n\)” leading to their \(n\).
or 75 855 from 0 or 1 up to 389
NB B1B1 can only be attained with consistent upper limits

Alternative method

M1: Correctly setting up \(S\) and writing it as the difference between two sums of squares.
or \(\displaystyle\sum_{r=1}^{780}r^2 - \left(2^2 + 4^2 + 6^2 + \ldots + 780^2\right)\)

B1: Correctly using the formula for the sum of squares up to and including 779 (or 780)
or 158 488 330 up to 780

M1: Bringing out a factor of \(2^2\) (or 4) to convert the sum of even squares to a sum of squares. Could be separate.

B1: Correctly using the formula for the sum of squares up to and including 389 (or 390)
or 19 849 115 up to 390
NB B1B1 can only be attained for 779 and 389 or 780 and 390

A1: \(158\,488\,330 - 4 \times 19\,849\,115\)

Notes

B1: AG. Writing the sum out so that the cancellation pattern is clear. If dots missing then some indication of existence of intermediate terms must be seen. Condone absence of + signs if listed by \(r\) value.
Must see \(r = 1\) and \(r = n\) and at least two justified cancellations.
Or algebraic cancellation: \(\displaystyle = \sum_{r=1}^{n}\frac{1}{r} - \sum_{r=1}^{n}\frac{1}{r + 1} = \sum_{r=1}^{n}\frac{1}{r} - \sum_{r=2}^{n+1}\frac{1}{r} = \frac{1}{1} + \sum_{r=2}^{n}\frac{1}{r} - \left(\sum_{r=2}^{n}\frac{1}{r} + \frac{1}{n + 1}\right) = 1 - \frac{1}{n + 1}\)

(Corrected from the printed mark scheme: the first line is printed with a stray bracket, as \(\ldots + \frac{1}{2} - \left(\frac{1}{3} + \frac{1}{3} - \frac{1}{4}\right) + \ldots\); the terms are \(-\frac{1}{3} + \frac{1}{3} - \frac{1}{4}\), as typed above.)

Notes

(b)(ii)

B1: For correct consideration of limit as \(n\) tends to infinity. Could be embedded but limit must be explicitly evaluated.
B0 for use of \(\infty\) in fraction (unless correct limit work seen) or incorrect limit notation eg \(\displaystyle\lim_{n \to \infty} \frac{1}{n + 1} \to 0\)

M1: Rewriting sum as difference of two sums with correct limits. Condone missing summation content but correct relationship between summation limits must be clear.

A1: If B0M0 then SCB1 for 0.01 from \(1 - 0.99 = 0.01\).

Notes

*M1: (1st) Correct PF expansion used in solution. Allow extraneous terms only if stated/evaluated as 0. Accept \(\frac{Ar + B}{r^2} + \frac{C}{r + 1}\) but not with additional \(\frac{D}{r}\) unless recovered later

*M1: (2nd) Recombining and use valid method to find coefficients (e.g. appropriate choice of \(r\) or comparing coefficients in \(r^2\), \(r\) or \(r^0\)). Indep of 1st M1 if from PF terms involving 3⁺ unknowns and their factors seen in a denominator

A1: (1st) Any one correct non-zero coefficient

A1: (2nd) Other two coefficients by valid method

dep*M1: Separating and expressing sum in form in which cancellation pattern is clear
\(r^{\prime} = r + 1\): \(\therefore \displaystyle\sum_{r=1}^{n}\frac{1}{r + 1} - \sum_{r=1}^{n}\frac{1}{r} = \sum_{r^{\prime}=2}^{n+1}\frac{1}{r^{\prime}} - \sum_{r=1}^{n}\frac{1}{r} = \frac{1}{n + 1} + \sum_{r=2}^{n}\left(\frac{1}{r} - \frac{1}{r}\right) - \frac{1}{1} = \frac{1}{n + 1} - 1\)

A1: (3rd) Complete argument with all detail. Allow embedded answers. Minimum of first and last cancellation terms shown (If algebraic approach, must see full argument)

Notes

M1: AG. Considering the limit as \(n\) tends to infinity of \(\frac{1}{n + 1}\) term; not \(\frac{1}{\text{infinity}}\)

A1: AG. Joined up argument. Some intermediate working must be shown. Condone poor, but clear limit notation

Notes

M1: Separating and using correct formulae for \(\sum r^3\) and \(\sum r^2\)

M1: Substituting must be seen

A1: Including all notation correct (Sigmas do not need limits)

Alternative method

M1: Separating and using the correct formulae

M1: Substituting anywhere in the algebra

Notes

M1: Correct form for partial fractions

A1: or \(A = 1,\ B = 3,\ C = -4\)

M1: Writing out sufficient terms so that cancellation pattern becomes evident. Could see \(N\) rather than 98.
If use \(r = a\) to \(N\) then M1 awarded when \(N = 98\) and \(N = k - 1\) both considered.

A1: Could see \(\dfrac{5k + 1}{k(k + 1)} - \dfrac{124}{2475}\)

M1: Equating their expression for LHS to 20539/34650 and rearranging to 3 term quadratic

A1: \(\dfrac{--61 \pm \sqrt{(-61)^2 - 4 \times 9 \times (-14)}}{2 \times 9} = \dfrac{61 \pm \sqrt{4225}}{18} = \dfrac{61 \pm 65}{18}\)
or \(9\left(\left(k - \dfrac{61}{18}\right)^2 - \left(\dfrac{61}{18}\right)^2\right) - 14 = 0\)
\(9\left(k - \dfrac{61}{18}\right)^2 - \dfrac{4225}{36} = 0\)

A1: Explicitly rejecting non-integer value, with reason, and deducing correct value… …or stating that the original function is undefined if \(r\) is allowed to take the value 0.
NB \(k \geqslant 0\) (or \(k \gt 0\)) alone is not a valid reason to reject \(-2/9\)

Notes

M1: Identifying the two series and quoting the standard series results

M1: Correctly cancelling to set up inequality with two layer fraction(s)
(NB M0 for \(\geqslant 342\)).
If cubic inequality formed then it must be factorised before M1…

A1: …and must be fully solved for A1.

A1: 512 from equation rather than inequality then M1M1A0A1

Notes

M1: partial fractions

M1: Allow any method to determine \(A\) and \(B\)

A1: Both values

M1: Use of differences

M1: Deal with subtraction

M1: Use of inequality on their formula

No marks for a purely numerical solution.

Notes

B1: Expanding and separating

M1: Use of formula for \(\displaystyle\sum_{r=1}^{100}r^2\)

Notes

B1: Correct factorisation of denominator soi

M1: Correct form for partial fractions

M1: Using partial fractions, separating into two sums, re-indexing so that the summands have identical form and cancelling central terms.
Might see start and end terms explicitly. eg
\(\displaystyle\sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{r=8}^{n+3} \frac{1}{r - 1} = \frac{1}{5 - 1} + \frac{1}{6 - 1} + \frac{1}{7 - 1} + \sum_{r=8}^{n} \frac{1}{r - 1} - \left(\sum_{r=8}^{n} \frac{1}{r - 1} + \frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2}\right)\)
Might see formal substitution of index. eg
Let \(R = r + 3 \Rightarrow r + 2 = R - 1\)
\(\therefore \displaystyle\sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{r=5}^{n} \frac{1}{r + 2} = \sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{R=8}^{n+3} \frac{1}{R - 1}\)

A1: AG.

Alternative method for last 2 marks

M1: At least these terms. M1 can be ft from any \(A\), \(B\) having opposite signs. For M1, condone omission of \(\dfrac{1}{7}\) or \(-\dfrac{1}{n - 1}\)

A1: AG. Correct cancellation to AG. Requires joined argument. Must have either clear diagonal cancellations with one explicit cancellation or less clear cancellation with numerical and algebraic cancellation shown or described

Notes

M1: Correct split of terms and use of formulae

A1: Correct forms for each summation. Can be earned even if 2 is dropped
\(\dfrac{1}{12}\left(3n^4 + 14n^3 + 21n^2 + 10n\right)\) earns 2 marks

A1: Fully factorised form for this mark

Notes

M1: Separate soi

M1: Use both formulae with \(n = 10\)

A1: AG. oe 1155 − 110

Alternative method (leaving substitution to the end)

Notes

B1: Correct factorisation of denominator soi

M1: Correct general form for partial fractions soi by correct answer. Could be incorrect if recovered. eg \(\frac{A}{r + 1} + \frac{B}{r + 2} + C\) and \(C = 0\)

A1: Both: \(\dfrac{1}{r + 1} - \dfrac{1}{r + 2}\)

M1: Condone omission of \(+\frac{1}{4}\) and \(-\frac{1}{n + 1}\) for M1 only

A1: AG. Cancellation must be evident

Notes

B1: Or \(\displaystyle\lim_{n \to \infty}\left(\frac{1}{2} - \frac{1}{n + 2}\right) = \frac{1}{2} - 0 = \frac{1}{2}\)

B1: Indication that \(1/(n + 2)\) is close to zero (accept “small”) when \(n\) is large

Notes

M1: Correctly combine both given fractions

Notes

M1: splitting fractions using above result

M1: method of differences attempted

A1: at least first three terms (for \(n = 2\), 3 and 4) shown. The last three for \(n - 2\), \(n - 1\) and \(n\) can be omitted, but if shown must be correct

B1*: or \(= 1 + \dfrac{1}{4}\) (if without the terms in \(n\))

B1: 0.3125 dep B1*

Notes

M1: converting series correctly to sigma notation

M1: writing their sum in terms of \(\sum_{r=1}^{n} r^2\) and \(\sum_{r=1}^{n} r\) soi

M1: substituting standard formulae for \(\sum_{r=1}^{n} r^2\) and \(\sum_{r=1}^{n} r\). \(r^2 + 2r\) or \(r(r + 2)\) must have been seen.

M1: correctly taking out a factor \(n\), \(n + 1\) or both from their expression; soi by correct final answer

A1: www but condone errors with sigma notation

Notes

B1: oe, \(\dfrac{1}{4}2n^2(2n + 1)^2\) is B0, mark final answer

Notes

M1: sum from \(n + 1\) to \(2n\) is sum from 1 to \(2n\) − sum from 1 to \(n\)

M1: taking out common factor of \(n^2\) (at any stage)

A1: or \(\dfrac{1}{4}n^2(15n^2 + 14n + 3)\) oe

Notes

B1: Denominator may be \(r^2 + 3r + 2\). Cannot be implied.

M1*: Enough correct terms to show cancellation in their series. First term must be correct. Fractions need not be simplified

A1: isw (or \(\frac{n}{2(n + 2)}\))

Alternative method

B1: Denominator may be \(r^2 + 3r + 2\). Cannot be implied.

M1*: Rewriting so both series have same fraction with correct limits

M1dep: Correct substitution of limits to leave two terms

A1: isw (or \(\frac{n}{2(n + 2)}\))

Notes

M1: (1st) splitting sum

A1 A1: \(\dfrac{1}{2}an(n + 1)\ldots\)
\(\ldots + bn\)

M1: (2nd) equating coefficients (or substituting two values for \(n\))

Alternative solution

M1A1: use of sum formula (need not be simplified)

M1: (3rd) equating coeffs or substituting two values for \(n\)

Alternative solution

As the question requires finding \(\displaystyle\sum_{r=1}^{n}(ar + b)\) in terms of \(a\), \(b\) and \(n\), allow a maximum of 3 marks for this method

Notes

M1: (2nd) enough terms to show consistent cancellation in their series

Notes

M1: For expanding correctly

Notes

M1: At least 2 terms of C + iS soi by correct GP formula

M1: sum to infinity of GP formula for their series (which must be geometric)

A1: oe

M1*: multiply numerator and denominator by \(3 - \mathrm{e}^{-2i\theta}\)

M1dep*: \(\mathrm{e}^{i\theta} = \cos\theta + i\sin\theta\) used when denominator has been simplified to a real expression

A1: AG

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

B1: Soi by M1A1

M1: Cannot use standard summation formulae for final 2 marks

A1: isw

Notes

M1*: \(\sum_{r=1}^{n} 1 = n\) used and starting to rearrange

M1dep*: Factorising with \(n\) or \(n + 1\) correctly

A1: Allow SC2 for correct solution using standard summation formulae.

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

M1: or cover-up method

Notes

M1dep*: showing cancellation clearly

SC: B3 For fully correct summation with A = \(-\frac{1}{2}\) and B = \(\frac{1}{2}\)

Notes

M1: \(r(r + 1)(r + 3)\)

M1: expanding and splitting sums

M1: substituting summation formulae

A1: correct expression

M1: drawing out \(n(n + 1)\)

Notes

M1: enough terms to show cancellation clearly
or \(\ldots + \dfrac{1}{n - 1} - \dfrac{1}{n + 1}\)
condone \(\ldots + \dfrac{1}{r - 1} - \dfrac{1}{r + 1}\)

A1cao: dep A1*

Notes

M1: \(\mathrm{e}^{\mathrm{i}\theta}.\mathrm{e}^{-\mathrm{i}\theta} = 1\)

M1: \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\) used

A1: NB AG

Notes

M1: at least 2 terms

A1: correct (\(n\)th term soi)

M1: sum of GP (condone \(S_\infty\))

A1: correct expression

M1: \(\times\) top and bottom by complex conjugate

M1: expand brackets; allow 1 slip, not on \(S_\infty\)

M1: taking real part; not on \(S_\infty\)

A1cao: NB AG; need evidence of clearing subsidiary denominators

Alternative solution

M1: substituting for cosines in terms of \(\mathrm{e}^{\mathrm{i}\theta}\)s; M1 A1: sum of GP; M1 A1: combining fractions; M1 A1: expanding; M1: expressing in cosines; A1: NB AG

(corrected from the printed mark scheme: in the printed alternative the first line has a stray factor \(\mathrm{e}^{\mathrm{i}\theta}\) after the first term and is cut off after \(\frac{1}{2^n}\); the second and third lines omit the overall factor \(\frac{1}{2}\); and the constant term in the fourth and fifth lines is printed as \(-2\) instead of \(-1\).)

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

B1: \(\frac{1}{3}n(n + 1)(2n + 1)\ldots\)

B1: \(\ldots - n\)

B1: factoring out \(n\) correctly

B1cao: allow \(\frac{2}{6}n(2n - 1)(n + 2)\), etc

Notes

M1: must see at least one cancellation

A1: or \(1 + \dfrac{1}{2} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\)

A1cao: must be to 4 DP

Notes

M1: combining fractions

A1: or \(\dfrac{2}{4r^2 - 1}\)
mark final answer

Notes

M1: (1st) \(r\)th term is \(\dfrac{2}{(2r - 1)(2r + 1)}\) soi

M1: (2nd) splitting into partial fractions soi

M1: (3rd) must end with \(\ldots - \dfrac{1}{2n + 1}\)
not \(\ldots - \dfrac{1}{2r + 1}\)

A1: \(1 - \dfrac{1}{2r + 1}\) is A0

M1: (4th) or \(\dfrac{2n}{2n + 1} \gt 0.999\,999\) allow equality, condone \(r\)
dep use of difference method [e.g. not by inspection]

M1: (5th) re-arranging (correctly) for \(n\)

A1cao: trial and error: must show both \(n = 499999\) and \(n = 500000\) for final A1