Notes

B1: Checks the closed form works for \(n = 1\) and \(n = 2\). Allow if they use the recurrence to find \(u_3\) and check for \(n = 2\) and \(n = 3\), but a consecutive pair must be checked.

M1: Makes the inductive assumption. If not explicitly made, accept just stating “\(n = k\) and \(n = k + 1\)” as making the assumption these are true – or implied by use of the relevant formulae, as long as the assumption is made clear in the conclusion. May use e.g. \(n = k - 2\) and \(n = k - 1\) instead and show true for \(n = k\). It must be clear it is the closed forms they are assuming, not a recurrence form.

M1: Substitutes expression for \(n = k\) and \(n = k + 1\) (or equivalents) into the recurrence formula.

M1: Uses algebra in an attempt to achieve the required result

e.g. Uses the coefficients of 6 and 8 to

• write as \(2^n\) terms as \(2^{k+2}\) and simplify.
• write as \(4^n\) terms as \(4^{k+1}\) or \(4^{k+2}\) and simplify.
  Note this is a method mark so you may score for the attempt even if some of the working is incorrect as long as the intent to reach the correct form is clear and at least one bit of indec work is correct.

A1: Completes the process correctly to the required form

A1: Correct conclusion made at the end. Depends on all three M’s and the A being gained and an attempt at both \(n = 1\) and \(n = 2\) having been shown true. Must convey the underlined ideas of

• true for \(n = 1\) and \(n = 2\)
• if true for two successive cases, it is also true for the next case
• a suitable conclusion that it is true for all positive \(n\)

though accept equivalent wordings for these.

Note Accept work with \(n\) instead of \(k\) throughout the inductive step.

Notes

M1: Sets closed form = 0 or < 0 and solves to set up an inequality, or to find a non-zero value, for \(2^n\). Alternatively if the \(4^n\) is not written in terms of \(2^n\) score for a correct process for taking logs to get a linear equation in \(n\). Be tolerant with incorrect inequalities for the M marks.

dM1: Solves \(2^n = a\) where \(a > 0\) by any valid means. May be by inspection. In the Alt it is for proceeding to a value for \(n\) from the linear equation.

A1: Deduces the first negative term of the sequence

Notes

B1: For explaining 2 of the 3 aspects as above. Allow attempts that convey the right idea even if not precisely described.

B1: All 3 aspects explained with sufficient detail shown. Must see the 1.0015 explained, not just “the 1.0015 is the 0.15%” or such.

Notes

B1: See scheme for answers. Must refer to the model, do not accept answer about “rounding” values.

Notes

M1: A complete method to solve the recurrence relation using \(B_n = \text{CF} + \text{PS} = a(1.0015)^n + b\)

M1: Uses \(\text{PS} = b \Rightarrow b = 1.0015b - 0.9\) to find a value for \(b\) (corrected from the printed mark scheme: the printed note has \(b = 1.0015b - 900\), but \(B_n\) is in thousands of pounds, so the constant is 0.9 as in the scheme)

A1: \(b = 600\)

M1: Uses \(B_0\) and their value for \(b\) to find a value for \(a\)

A1: Fully correctly defined sequence \(B_n = 600 - 420(1.0015)^n \quad (n \geqslant 0)\)

Notes

M1: Uses a correct strategy to identify the value for \(n\). NB “impossible equations” score M0.

A1: Correct number of years and months.

Notes

B1: For explaining any two of the three aspects in the scheme.

B1: All three aspects explained.

Notes

M1: Sets up and solves the auxiliary equation.

A1: Correct complementary part found.

M1: Selects correct form for particular solution and substitutes and combines result with their CF

M1: Uses the initial value to find the constant.

A1: Correct solution.

Notes

M1: Attempts a scale factor with their \(u_8\). Accept attempts at scaling by \(25 \times 3^{-k}\) or \(25 \times 9^{-k}\) where \(k\) is 7, 8 or 9.

A1: Correct answer.

Notes

M1: Forms and solves the auxiliary equation.

A1: Correct single root of 3.

M1: Selects the correct form for \(u_n\) for their roots of the equation. (If distinct real roots were found allow for \(u_n = A\alpha^n + B\beta^n\))

M1: Uses the values of \(u_1\) and \(u_2\) with the corresponding values of \(n\) used to form and uses a correct method to solve simultaneous equations to find the constants. If no method is shown, use of calculator, for solving the simultaneous equations the values must be correct for their equations.

A1: Correct closed form, isw

Notes

M1: Forms and solves the auxiliary equation.

A1: Correct roots – either Cartesian or polar form, award when first seen and isw.

A1ft: Correct complementary function, follow through on their first complex roots. (so A1A0 if roots initially correct but error simplifying leads to wrong CF). Note: use of power \(n + 1\) or \(n - 1\) in Cartesian form is also fine

M1: Correct form for the particular solution and a complete method to find the PS.

A1: Correct general solution, either form

M1. Substitutes \(n = 1\) and sets equal to 4 and substitutes \(n = 0\) and sets equal to 1. To find the values of the constants

A1: Correct solution, any form.

Notes

B1: Uses the model to state \(A = 50\)

B1: A correct explanation

Notes

M1: A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda\)

M1: Uses their value of \(A\) with \(\text{PS} = \lambda \Rightarrow \lambda = 1.005\lambda - \text{“}50\text{”}\) to find a value for \(\lambda\)

A1: \(\lambda = 10000\)

M1: Uses \(U_0 = 1000\) and their value of \(\lambda\) to find the value of \(c\)

A1: Fully correctly defined sequence \(U_n = 10000 - 9000(1.005)^n\)

Alternative

M1: A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda\)

M1: Uses \(U_0\) to form an equation for their \(c\) and their \(\lambda\)

M1: Uses the given formula to find \(U_1\) and then uses their value to form another equation for their \(c\) and their \(\lambda\)

A1: Correct second equation using \(U_1\)

A1: Solve simultaneously to find \(U_n = 10000 - 9000(1.005)^n\)

Notes

M1: Sets their expression = 0 and solves using logarithms to find a value for \(n\).

A1: 22 (months)

Note: Using trail and error must show that \(U_{21} \gt 0\) is positive and \(U_{22} \lt 0\)

Notes

M1: See scheme. Explains how the assumptions lead to at least two of the aspects indicated. Accept less formal explanations as long as the intent is clear.

A1: All three aspects explained and put together to set up the model.

Notes

B1: Checks the case for \(n = 1\) holds.

M1: Makes the inductive assumption (may be implied by working) and substitutes the closed form for \(U_k\) into the recurrence relation for \(U_{k+1}\) or equivalent work with different variable (e.g. \(n\) instead of \(k\)) or indexing (e.g. from \(k - 1\) to \(k\)).

M1: Simplifies to the point of combining the powers of \(\dfrac{4}{5}\) to one term.

A1: For correct work leading to the form shown. The \(k + 1\) must be seen in the added term but allow just \(k\) for the power.

A1: For a completely correct proof (all previous marks must be gained) with a conclusion that includes all of the bold statements in the scheme or equivalents.

Alternative

B1: Checks the case for \(n = 1\) holds.

M1: Writes out the term \(U_{k+1}\) and starts the process to write in terms of \(U_k\) by factorising out \(\dfrac{4}{5}\) from the first term.

M1: Factorises out \(\dfrac{4}{5}\) to form \(= \dfrac{4}{5} \times \left(325\left(\dfrac{4}{5}\right)^{k-1} + 100k - 400\right) + \ldots\)

A1: For correct work leading \(U_{k+1} = 0.8U_k + 20(k + 1)\)

A1: For a completely correct proof (all previous marks must be gained) with a conclusion that includes all of the bold statements in the scheme or equivalents.

Notes

M1: Evaluates \(U_{24}\) \(U_{25}\) \(U_{26}\) or \(U_{27}\). Allow attempts that deduce \(\left(\dfrac{4}{5}\right)^{n} \to 0\) and just evaluate \(100 \times 26 - 400 = 2200\)

A1: Correct value and appropriate conclusion.

Notes

M1: Forms and solves the auxiliary equation.

A1: Correct complementary function.

M1: Correct form for the particular solution, substitutes into the recurrence relation to find the PS.

A1: Correct general solution

(corrected from the printed mark scheme: the first line is printed in a garbled symbol font)

Notes

Note: They must have two constants to score any marks in this part

M1: A complete method to find the constants using the information given. Form two equations and solves simultaneously to find values for \(A\) and \(B\).

A1: Correct solution.

Notes

B1: For explaining any two of the three aspects in the scheme.

B1: All three aspects explained.

Notes

B1: For \(u_1 = 18\) found in order to be able to form and solve simultaneous equations.

M1: Setting up and solving equations for \(a\) and \(b\) using their \(u_1\) and 20, with at least one value found.

A1: Either \(a\) or \(b\) correct.

A1: Both \(a\) and \(b\) correct.

Alt (b)

B1: Uses \(n = 0\) to form a correct relation relating \(a\) and \(b\).

M1: Substitutes the given general form into the recurrence relation and extracts an equation in \(b\) only.

A1: Correct value for \(b\)

A1: Both \(a\) and \(b\) correct.

Alt 2 (b)

B1: Uses particular solution is constant to find \(b\)

M1: Substitutes for \(n = 0\) or \(n = 1\) and sets equal to 20 in the general form to form an equation in \(a\)

A1: Correct equation follow through their \(b\).

A1: \(a\) correct.

Notes

M1: Uses the model to work out the long-term behaviour or the amount of vitamin in the person. Accept identifying \(\dfrac{50}{3}\) as a lower limit for \(u_n\).

M1: Finds the lower bound for the amount of vitamin in the person, subtracting 10 from their long-term value or multiplying the \(\dfrac{50}{3}\) limit by 0.4 to find amount just before next tablet.

A1: Concludes the course of vitamin will be effective following correct work and reasoning.

NB: Attempts that set \(u_n = 6\) and try to solve for \(n\) score no marks. Attempts that set \(u_n - 10 = 6\) and try to solve for \(n\) can score the second M only. Long term behaviour needs to be considered for first M.

Notes

B1: Explains has to move from A

Notes

B1: Shows the statement is true for \(n = 1\). Needs to show that \(p_1 = 0\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)

M1: Makes a 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: Finds an expression for \(p_{k+1}\) using the recurrence relation and substitutes in for \(p_k\)

M1: Shows a correct intermediate stage \(p_{k+1} = \frac{1}{3} + \frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k\)

Condone \(\frac{1}{3} - \frac{2}{3}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k\)

A1: \(p_{k+1} = \frac{2}{3}\left(-\frac{1}{2}\right)^{k+1} + \left(\frac{1}{3}\right)\) cso

A1: Correct complete conclusion. This mark is dependent on previous four marks, the B mark is not required. 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.

Note Using \(p_{k+1} = \frac{1}{2} - \frac{1}{2}p_k = \frac{1}{2} - \frac{1}{2}\left[\frac{2}{3}\left(-\frac{1}{2}\right)^k + \left(\frac{1}{3}\right)\right] = \frac{1}{2} + \frac{2}{3}\left(-\frac{1}{2}\right)^{k+1} - \left(\frac{1}{6}\right)\) scores 2^nd M1 3^rd M1

Notes

B1: see scheme

Notes

B1: Checks the closed form works for \(n = 1\) and \(n = 2\)

M1: Makes the inductive assumption. May use e.g. \(n = k - 2\) and \(n = k - 1\) instead and show true for \(n = k\). It must be clear it is the closed forms they are assuming, not a recurrence form.

M1: Substitutes expression for \(n = k\) and \(n = k + 1\) (or equivalents) into the recurrence formula.

M1: Takes out common factors of at least \((-3)^k k!\) in their expression, or equivalent for their assumed true values. Treatment of the \((-3)\) must be correct, but condone invisible brackets if recovered.

Note: they may well take out more at this stage, which is fine, e.g.
\(u_{k+2} = 9(k+1)^2\left((-3)^k k!\right) - 3\left((-3)^{k+1}(k+1)!\right) = (-3)^{k+2}(k+1)!\left[(k+1) + 1\right]\)

A1: Simplifies correctly to the required form for their assumed true values.

A1: Correct conclusion made. Depends on all three M’s and the A being gained. Must convey the ideas of 1) true for \(n = 1\) and \(n = 2\), 2) if true for two successive cases, it is also true for the next case and 3) a suitable conclusion that it is true for all positive \(n\).

Notes

B1: Correct equation

Notes

B1: Demonstrates that the result is true for \(n = 1\)

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

A1ft: Correct expression for \(A_{k+1}\)

A1: Reaches the correct statement for \(k + 1\) in the required form

A1: Completes the inductive argument

Notes

M1: Identifies a correct strategy using \(n = 16\) to obtain an equation in \(F\)

M1: Proceeds to obtain a value for the minimum value of \(F\)

A1: Correct numerical expression for \(F\)

A1: Correct answer (allow to the nearest penny or nearest £)

Notes

M1: Forms and solves the auxiliary equation.

A1: Correct (repeated) root found.

M1: Forms the correct complementary function for their (real) root(s) to the equation, \((A + Bn)r^n\) if repeated root, or allow \(Ar_1^{\,n} + Br_2^{\,n}\) if distinct real roots are found.

M1: Attempts to use a particular solution of the correct form (ie \(an + b\) or a higher order polynomial in \(n\) containing this) in the recurrence relation.

dM1: Expands and solves for \(a\) and \(b\)

A1: Correct values for \(a\) and \(b\)

B1ft: Forms the general solution as the sum of their complementary function and a particular solution of correct form with their \(a\) and \(b\)

M1: Applies the initial values and solves for the constants

A1: Correct answer.

Notes

B1: Need to explain that 2% interest rate linked to multiplication by scale factor 1.02

B1: Need to explain that 500 is added due to receiving £500 each year

B1: Needs to explain that \(U_0 = 1000\) is the initial amount invested

Notes

B1: See main scheme

Notes

M1: A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.02)^n + \lambda\)

M1: Uses \(\text{PS} = \lambda \quad \Rightarrow \lambda = 1.02\lambda + 500\) to find a value for \(\lambda\)

A1: \(\lambda = -25\,000\)

M1: Uses \(U_0\) and their value for \(\lambda\) to find a value of \(c\)

A1: Fully correctly defined sequence \(U_n = 26000(1.02)^n - 25\,000, \quad (n \geqslant 0)\)

Alternative 1

M1: A correct form for \(U_n\) term of a GP + Sum of a GP both with \(r = 1.02\)

M1: For the sum of a GP with \(a = 500\), \(r = 1.02\) and uses \(n\) or \(n - 1\)

A1: Correct the sum of a GP with \(a = 500\), \(r = 1.02\) and \(n\)

M1: For the term of a GP with \(a = 1000\), \(r = 1.02\) and uses \(n\) or \(n - 1\)

A1: Fully correctly defined sequence \(U_n\)

Notes

M1: Uses their \(U_n\) with either \(n = 7\) or 8

A1ft: Finds \(U_7\) compares with 4 500 and comes to an appropriate conclusion. Follow through on their value of \(U_7\)

Notes

B1: A correct expression for the model using the information given

Notes

M1: Forms and solves the auxiliary equation for their answer to (a) with \(k = 0.24\)

A1: The correct closed form deduced from their solutions. This must be consistent with their equation. Note the answer is given so check carefully. This is not a follow through mark.

B1ft: Applies initial conditions to their general equation – correct two equations for their general form with \(V_1 = 65\) and \(V_2 = 71\)

M1: Attempts to solve their equations showing a correct method, reaching a value for at least one variable. It is a show that question and answers are on the paper, so method is needed. Look for one equation multiplied through to give same coefficients before attempting eliminating or substitution. If a matrix system is used the inverse must be found, not just solutions stated.

A1*: Correct expression formed following suitable working with no errors seen. With fractions instead of decimals is fine.

Alternative (b): \(V_n = 50(1.2)^n - 25(-0.2)^n\), \(V_1 = 65\), \(V_2 = 71\)

M1: Substitutes into equation for \(n = 1\) and \(n = 2\) to verify true for these cases.

A1: Deduces true for base cases and makes a correct assumption statement. This must include two successive cases assumed true, so e.g. as in scheme, or with \(k - 2\) and \(k - 1\) etc, or may assume true for all (integers) \(k \leqslant n\). But do not allow if assumed true for just \(k\).

B1ft: Substitutes the formula for \(k\) and \(k + 1\) (or their successive values) into the recurrence formula, follow through their equation from part (a).

M1: Rearranges to the form \(a(1.2)^{(k+2)} + b(-0.2)^{k+2}\)

A1*: Correct work leading to the correct equation for \(V_{k+2}\) and makes suitable inductive conclusion, including the ideas of “true for \(n = 1\) and \(n = 2\)”, “if true for \(n = k\) and \(n = k + 1\) then true for \(n = k + 2\)” and “hence true for all integers”.

Notes

M1: Selects a suitable method to solve the problem. For example, realises that in the model, \((-0.2)^n\) is negligible for large \(n\) and so attempts to solve e.g. \(50(1.2)^N = 10^6\), or tries at least one value either side of \(N = 55\) as a process of trial and improvement, or uses a calculator/graphical approach – implied by a value of \(N = 55\) or \(N = 54\) stated.

A1: \(N = 55\).

The correct answer will imply both marks for this part. Ignore erroneous working if correct answer is stated as a restart.

Notes

B1: Need to see 3% decrease linked to scale factor of 0.97

B1: Need to see that adding 0.15 corresponds to the yearly growth in metres. There must be some reference to the units for this mark.

B1: An explanation that \(H_1\) is the first term (the starting height) and this is 6m

Notes

B1: Begins proof by induction by considering \(n = 1\) and obtains \(H_1 = 6\)

M1: Assumes true for \(n = k\) and uses iterative formula to consider \(n = k + 1\)

A1: Reaches \((0.97)^k + 5\) with no errors

B1: Correct conclusion. This mark is dependent on all previous marks apart from the first 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

B1: States the height will approach 5m

Notes

M1: Uses the model to adopt a correct strategy to find the required percentage

A1: Interprets their answer correctly in terms of the original context