Notes

B1: CAO for complementary function (condone if stated as \(u_n = A\alpha^n\))

M1: correct form for particular solution and substituted into recurrence relation

dM1: simplifying (by eliminating \(n\)) and solving for their constant (\(B\))

A1: correct general solution (must be \(u_n =\) ) (must be fully simplified)

Notes

M1: Form two equations using their general solution, and both given conditions

dM1: Eliminate either \(A\) or \(\beta\) from their two equations (formed from the given conditions)

ddM1: Re-arrange to obtain an expression for either \(\beta\) or \(A\) in terms of \(L\), \(k\), \(T\), \(n\) and \(\alpha\)

They may eliminate \(A\) and then obtain an expression for \(\dfrac{\beta}{\alpha - k}\) and then substitute for this

dddM1: Eliminate both \(A\) and \(\beta\) from their general solution to obtain an expression for \(u_n\) in terms of \(L\), \(k\), \(T\), \(n\) and \(\alpha\) only

A1: CSO – correctly derive given result (so sufficient working must be shown)

In part (b) for those that at attempt to verify (rather than show the given result) then the first 3 M marks are as in the main scheme (M1 for using the given conditions to form two equations, dM1 for eliminating \(A\), ddM1 for obtaining an expression for \(\beta\) (which for reference if correct is \(\beta = \dfrac{L\alpha^T(\alpha - k)}{\alpha^T - k^T}\)), then the fourth M mark is for obtaining an expression for \(u_{n+1} - \alpha u_n\) in terms of \(L\), \(k\), \(T\), \(n\) and \(\alpha\) only – which if correct should be equivalent to \(u_{n+1} - \alpha u_n = \dfrac{Lk^n}{1 - \left(\frac{k}{\alpha}\right)^T}\left\{k - k\left(\frac{k}{\alpha}\right)^{T-n}\left(\dfrac{\alpha}{k}\right) - \alpha + \alpha\left(\frac{k}{\alpha}\right)^{T-n}\right\} = -k^n\left\{\dfrac{L(\alpha - k)}{1 - \left(\frac{k}{\alpha}\right)^T}\right\}\). The final A mark is for correctly showing that \(\dfrac{L(\alpha - k)}{1 - \left(\frac{k}{\alpha}\right)^T}\) is equivalent to \(\dfrac{L\alpha^T(\alpha - k)}{\alpha^T - k^T}\) (so explicitly showing that \(u_{n+1} - \alpha u_n\) is equal to \(-\beta(k^n)\) with a correct expression for \(\beta\) explicitly seen (in terms of \(L\), \(k\), \(T\) and \(\alpha\))). (corrected from the printed mark scheme: the printed note says “In part (c)”; it refers to part (b))

Alternative for (b)

If they eliminate \(A\) first

\(\beta = \dfrac{L\alpha^T(\alpha - k)}{(\alpha^T - k^T)}\) and \(A = L - \dfrac{L\alpha^T}{(\alpha^T - k^T)} = \dfrac{-Lk^T}{\alpha^T - k^T}\)

Notes

(i) B1: CAO for \(\alpha\)

(ii) B1: CAO for \(k\)

(iii) B1: CAO for the relevance of \(\beta\) in (*) must be clear that this is the amount repaid one year after the loan is taken out. Accept e.g original amount repaid or initial repayment and condone reference to 0^th year if clear that this is the first repayment (do not accept annual repayment)

Notes

M1: Substitute \(T = 40\), \(n = 30\) and their values of \(\alpha, k\) into correct particular solution (accept sight of \(1.05L\) for this mark even if full calculation not shown)

A1: CAO that the outstanding debt after 30 years is larger than the original debt (so must see some reference to \(1.05L\) (or better) and \(L\)) (accept \(1.05L > L\)) (must follow from a correct calculation)

Notes

a1B1: CAO

a1M1: Correct linear form for particular solution and substituted into recurrence relation.

a1A1: CAO

a2A1ft: General solution correct or ft their values for \(a\) and \(b\) and/or C.F.

a1M1: Correct substitution.

a1A1: \(u_n =\) fully correct expression .

Notes

b1B1: Even terms \(\gt 0\)

b1M1: Odd terms after \(u_1 \geqslant 0\)

b1A1: CAO

Notes

Note mark (a) and (b) together

M1: correct auxiliary equation and attempt to solve (leading to two distinct values of \(m\))

A1: CAO for complementary function (accept if seen in subsequent working)

M1: correct form for particular solution and substituted \(n + 2,\ n + 1,\ n\) into recurrence relation

dM1: compares coefficients and setting up both equations in \(a, b\) (so one equation in \(a\) only and one equation in \(a\) and \(b\) (although \(a\) may have already been found from the first equation)) – dependent on previous M mark (\(a = 2\) and \(b = -4\))

A1ft: correct general solution following through their complementary function (must be their C.F. + \(2n - 4\)) Please award for a fully correct solution seen either here or being used in (b)

Notes

ddM1: use correct initial conditions correctly to form simultaneous equations in their \(A\), \(B\) and \(k\) and attempt to solve for \(A\) and \(B\) – dependent on the previous two M marks. Award this mark for Either

• eliminating either \(A\) or \(B\) from the correct two equations leading to one of \(7A = 2k + 34\)   or   \(7B = 1 - 2k\)   (accept any equivalent form)

Or

• eliminating both \(A\) and \(B\) from their equations using the initial conditions and obtaining expressions in terms of \(k\)

If correct \(A = \dfrac{34 + 2k}{7},\ B = \dfrac{1 - 2k}{7}\)

dddM1: setting coefficient (which must be a linear expression in \(k\)) of their \((-3)^n\) equal to zero – dependent on the previous three M marks (and at least one root of the auxiliary equation being negative)

A1: CAO (must be from correct working)

Notes

B1: CAO

Notes

B1: CAO

M1: substituting their trial solution into the recurrence relation in an attempt to find their \(\lambda\)

A1: CAO

M1: using the conditions in the model to calculate \(A\)

A1: CAO (must be \(u_n =\) not \(u_{n+1} =\))

Alternative

General solution is \(u_n = A(1.025)^{n-1} - 72000\)
\(u_n = 78925(1.025)^{n-1} - 72000\)

Notes

B1ft: substitutes \(n = 0\) in their solution to find \(D\)

Alternative

\(6925 = 1.025D + 1800 \quad \Rightarrow D = 5000\)

Notes

M1: sets their particular solution greater than 20000– their particular solution must be of the correct form \(\left(u_n = c(1.025)^n \pm d \text{ or } u_n = c(1.025)^{n-1} \pm d\right)\)

M1: dependent on previous M mark – correctly re-arranges and then applies the process of taking logs for their particular solution

A1: CAO

Alternative

\(78925(1.025)^{n-1} - 72000 \gt 20000\)
\((1.025)^{n-1} \gt \dfrac{3680}{3157} \Rightarrow (n-1)\log(1.025) \gt \log\left(\dfrac{3680}{3157}\right)\)
\(n - 1 \gt 6.20795\ldots \Rightarrow n = 8\)

Notes

B1: any correct expression for \(u_{n+1}\)

M1: eliminating \(A\) to form a first order recurrence relation containing \(u_{n+1}\) and \(u_n\)

A1: CAO for \(u_{n+1} - 3u_n = -10n^2 + 10n + 3\) (need not explicitly state \(a = -3, b = -10, c = 10, d = 3\))

Alternative 1

B1: Considers the C.F. and deduces \(a = -3\) with no other values stated

M1: Obtains P.S. and forms equation using \(u_{n+1} \pm 3u_n\)

A1: CAO for \(u_{n+1} - 3u_n = -10n^2 + 10n + 3\) (need not explicitly state \(a = -3, b = -10, c = 10, d = 3\))

Alternative 2

B1: Forms the correct equation for \(u_{n+1} + au_n\)

M1: Attempts to compare coefficients – at least three terms seen

A1: CAO

Notes

B1: CAO

B1: CAO

M1: substituting their trial solution into the recurrence relation (which must be of the form \(u_{n+1} = \alpha u_n + k\) where \(\alpha = 0.75\) or \(\pm 0.25\)) in an attempt to find their \(\lambda\)

A1: CAO for the general solution (may be implied by subsequent working)

M1: using the conditions in the model and their expression of the form \(A(\alpha)^n + Bk\) where \(\alpha\) is 0.75 or \(\pm 0.25\) to calculate \(A\) (which if correct is \(-4k\))

A1: CAO for the particular solution (may be an expression without \(u_n =\) accept any equivalent form e.g. \(u_n = 4k - 3k(0.75)^{n-1}\))

Notes

dM1: Applying \(n = 8\) to their general solution of the form \(A(\alpha)^n + Bk\) where \(\alpha\) is 0.75 or \(\pm 0.25\) equal to 1750 (dependent on both M marks in (a))

A1: CAO – must be 486.18

Notes

M1: Substituting \(u_n = v_n + 2n\) to obtain a second-order recurrence relation in \(v_n\) only

A1: CAO Correct homogeneous second-order recurrence relation

dM1: Solving their three-term auxiliary equation (dependent on previous M mark) (if no auxiliary equation seen, correct expression implies this mark)

A1ft: Correct general solution for \(v_n\) following their roots or complimentary function if they do not obtain homogeneous relationship.

B1: Either converting the first two given values for the sequence for \(u\) to \(v\) or for stating the corresponding general solution for \(u_n\)

M1: Setting up two equations using their solution and solving for their \(A\) and \(B\) (if correct \(A = 17\) and \(B = -2\)) (dependent on general solution of the form \(A + B(\alpha)^n\) where \(\alpha\) is \(-\dfrac{1}{2}\) or \(-2\))

A1: CAO (may be an expression without \(u_n =\) accept any equivalent form e.g. \(u_n = 17 + \left(-\dfrac{1}{2}\right)^{n-1} + 2n\))

Special Case – does not use the transformation Max 6/8

M0: Transformation not used

A0

M1: Setting up and solving their three-term auxiliary equation (if no auxiliary equation seen, correct expression implies this mark)

A1ft: Correct complementary function for \(u_n\) following their roots

B1: Finding the correct particular solution and stating the corresponding general solution for \(u_n\)

dM1: Setting up two equations using their solution and solving for their \(A\) and \(B\) (if correct \(A = 17\) and \(B = -2\)) dependent on previous M mark (dependent on general solution of the form \(A + B(\alpha)^n + Cn\) where \(\alpha\) is \(-\dfrac{1}{2}\) or \(-2\))

A1: CAO (may be an expression without \(u_n =\) accept any equivalent form e.g. \(u_n = 17 + \left(-\dfrac{1}{2}\right)^{n-1} + 2n\))

(b) B1: Explanation that as \(n\) becomes large the terms of the sequence are (approximately) in arithmetic progression (or equivalent) dependent on correct expression for \(u_n\)

Notes

B1: Explanation that as \(n\) becomes large the terms of the sequence are (approximately) in arithmetic progression (or equivalent) dependent on correct expression for \(u_n\)

Notes

B1: cao (or equivalent e.g. \(A(1.5)^{n-1}\))

M1: correct form for particular solution e.g. \(\alpha n^2 + \beta n + \gamma,\ \alpha(n-1)^2 + \beta(n-1) + \gamma,\) etc. (so anything that is equivalent to a three term quadratic in \(n\) with unknown coefficients and constant term) together with a valid substitution into recurrence relation (so not substituting \(u_n\) on both sides of the recurrence relation)

dM1: compares coefficients and setting up all three equations in \(\alpha, \beta, \gamma\) - dependent on previous M mark

A1: cao for the values of \(\alpha, \beta, \gamma\)

ddM1: use correct initial condition correctly to form a linear equation in their \(A\) and \(k\) – dependent on the two previous M marks

A1: correct particular solution (in terms of \(k\)) – need not see \(u_n =\) but if seen then it must be correct (and therefore \(u_{n+1} =\) is A0)

Notes

M1: dependent on all M marks in (a) – substituting \(n = 10\) and setting the particular solution > 5 000 (or equal to)

A1: cao (125) from correct working so must have had a correct expression for \(u_n\) in (a) (so dependent on at least the first 5 marks in (a))

Additional guidance:

Those candidates who re-write \(u_{n+1} = \frac{3}{2}u_n - 2n^2 - 4\) as \(u_n = \frac{3}{2}u_{n-1} - 2(n-1)^2 - 4\) can score full marks. Their solution will usually begin: CF is \(A(1.5)^n\) then a PS of the form \(\alpha n^2 + \beta n + \gamma\) leading to

and then \(\frac{3}{2}\alpha - 2 = \alpha \Rightarrow \alpha = 4,\ -3\alpha + \frac{3}{2}\beta + 4 = \beta \Rightarrow \beta = 16,\ \frac{3}{2}\alpha - \frac{3}{2}\beta + \frac{3}{2}\gamma - 6 = \gamma \Rightarrow \gamma = 48\) and then their general solution should be as in the main scheme (although for the final mark in (a) do look out for those who call the left-hand side \(u_{n+1}\))

It is common for candidates to re-write \(u_{n+1} = \frac{3}{2}u_n - 2n^2 - 4\) as \(u_n = \frac{3}{2}u_{n-1} - 2n^2 - 4\) which is incorrect. This can score all B and M marks in both parts only.

Slightly less common is to have a CF of the form \(A(1.5)^{n-1}\) and a PS of the form \(\alpha(n-1)^2 + \beta(n-1) + \gamma\) this leads to \(u_{n+1} = A(1.5)^{n-1} + 4(n-1)^2 + 32(n-1) + 96\)
Now to use the initial condition correctly \(u_0 = k \Rightarrow u_1 = \frac{3}{2}k - 4\) (oe) and this leads to \(A = \frac{9}{4}k - 108\)
So, \(u_{n+1} = \left(\frac{9}{4}k - 108\right)(1.5)^{n-1} + 4(n-1)^2 + 32(n-1) + 96\) which when re-written in terms of \(u_n\) gives the form as in the main mark scheme

Of course, any CF of the form \(A(1.5)^{n \pm k_1}\) and any PS of the form \(a(n \pm k_2)^2 + b(n \pm k_3) + c\) where \(k_1, k_2, k_3\) are constants will work. So, award the M marks for the correct methods as illustrated in the notes in mark scheme and the first A mark in (a) for three correct coefficients of their PS and the second A mark in (a) for a fully correct expression

Notes

B1: CAO

B1: CAO

M1: substituting their trial solution into the recurrence relation in an attempt to find their \(\lambda\) (which if correct is \(\dfrac{k}{1-p}\))

A1: CAO for the general solution

M1: using the conditions in the model to calculate \(A\) (which if correct is \(p^{-1}\left(5000 - \dfrac{k}{1-p}\right)\))

A1: CAO for the particular solution (oe)

Notes

B1: Applying \(u_m \geqslant 135000\) (or equality or strict inequality) to their general solution together with correct values for \(k\) and \(p\) (dependent on both M marks in (a))

M1: dependent on previous B mark – solving their equation using logarithms

A1: CAO – must be rounded to 23

Special case: A common misread is 500 for 5000. Mark as a misread so final A marks in (a) and (b) deducted.

(corrected from the printed mark scheme: the inequality signs \(\geqslant\) and \(\leqslant\) in this part are printed as “…” and “,,” because of a font error)

Notes

B1: cao (or equivalent e.g. \(A(-3)^{n-1}\)) – condone \(A(-3^n)\) in (a) for all but the final A mark

M1: correct form for particular solution e.g. \(an + b,\ a(n-1) + b,\) etc. (so anything that is a constant times \(n\) + a constant) and substituted into recurrence relation

dM1: compares coefficients and setting up both equations in \(a, b, k\) (so one equation in \(a\) only and one equation in \(a\), \(b\) and \(k\) (although \(a\) may have already been found from the first equation)) – dependent on previous M mark

ddM1: solve for \(a\) and \(b\) (with \(b\) in terms of \(k\)) – dependent on both previous M marks

A1: a correct general solution (in terms of \(k\)) – ignore labelling of left-hand side

dddM1: use correct initial condition correctly to form an equation in their \(A\) and \(k\) – dependent on all three previous M marks

A1: correct particular solution (in terms of \(k\)) – must have correct left-hand side

Notes

M1: Setting the coefficient of the exponential term equal to zero and solving for \(k\) – their solution from (a) must be of the form \(\alpha(\beta)^n + \gamma n + \delta\) (where \(\alpha, \beta, \gamma, \delta\) are constants and \(\alpha, \delta\) are in terms of \(k\) only)

dM1: Substituting their value of \(k\) to obtain an expression for \(u_n\) which is linear and substituting \(n = 100\) (dependent on previous M mark)

A1: cao (26) – from correct working including a correct expression for \(u_n\) in (a)

Additional guidance:

Those candidates who re-write \(u_{n+1} + 3u_n = n + k\) as \(u_n + 3u_{n-1} = (n-1) + k\) can score full marks. Their solution will usually begin: CF is \(A(-3)^n\) then a PS of the form \(an + b\) leading to \(4an + 4b - 3a - n + 1 - k = 0\) and so \(a = \frac{1}{4}, b = \frac{4k-1}{16}\) and then their general solution should be as in the main scheme (although for the final mark in (a) do look out for those who call the left-hand side \(u_{n+1}\))

It is common for candidates to re-write \(u_{n+1} + 3u_n = n + k\) as \(u_n + 3u_{n-1} = n + k\) which is incorrect. This can score all B and M marks in both parts only

Slightly less common is to have a CF of the form \(A(-3)^{n-1}\) and a PS of the form \(a(n-1) + b\) this leads to
\(a(n-1) + a + 3(a(n-2) + b) = n + k\) so \(4an - 7a + 4b = n + k\) therefore \(a = \frac{1}{4}\) and \(b = \frac{4k+7}{16}\)
So, giving as a general solution \((u_{n+1} =)\,A(-3)^{n-1} + \frac{1}{4}(n-1) + \frac{4k+7}{16}\)
Now to use the initial condition correctly \(u_0 = 1 \Rightarrow u_1 = k - 3\) (oe) and this leads to \(A = \frac{153 - 36k}{16}\)
So, \(u_{n+1} = \left(\frac{153 - 36k}{16}\right)(-3)^{n-1} + \frac{1}{4}(n-1) + \frac{4k+7}{16}\) which when re-written in terms of \(u_n\) gives the form as in the main mark scheme

Of course, any CF of the form \(A(-3)^{n \pm k_1}\) and any PS of the form \(a(n \pm k_2) + b\) where \(k_1, k_2\) are constants will work. So, award the M marks for the correct methods as illustrated in the notes in mark scheme and the first A mark in (a) for a correct general solution (ignoring the labelling of the left-hand side) and the second A mark in (a) for a fully correct expression with correct left-hand side

Notes

M1: Correct auxiliary equation (may be implied by 6, 9 correct) from \(u_n = (A + Bn)(-3)^n\)

A1: CAO – allow values stated implicitly i.e., \(u_{n+2} + 6u_{n+1} + 9u_n = 0\)

Notes

M1: Uses \(u_0 = u_1 = 1\) and attempts to find \(A\) and \(B\)

A1: CAO – allow values stated implicitly e.g. \(u_n = \left(1 - \dfrac{4}{3}n\right)(-3)^n\)

Notes

B1: Correct reasoning to derive given result – sufficient working must be shown as recurrence equation given in question

Notes

B1: cao for auxiliary equation and corresponding solutions (this mark can be implied by the correct complementary function)

B1: cao for the complementary function

M1: Substitute \(x_n = \lambda\) into their second order recurrence relation and solve for \(\lambda\)

A1: Correct general solution

M1: Forms one equation in \(A\) and \(B\) using \(x_1 = 1\)

B1: Uses original recurrence relation for \(x_{n+1}\) to derive the expression \(2a + 3\)

M1: Setting up a second equation in \(A\) and \(B\)

A1: cao (oe e.g., \(x_n = 0.25(a+1)(4)^n - 0.5(a+1)(2)^n + 1\))

Notes

M1: Using \(x_7 = 28225\) to form a linear equation in \(a\) and attempt to solve for \(a\)

A1: cao for \(a\)

Notes

M1: Uses given complementary function to find auxiliary equation corresponding to second-order recurrence relation

A1: cao for both \(\alpha\) and \(\beta\)

Notes

M1: substitute \(u_n = \lambda(-3)^n\) into their second-order recurrence relation

M1: forms linear equation in \(\lambda\) only

A1: correct general solution

M1: use \(2u_0 = u_1\) to form an equation in \(B\) (and possibly \(A\))

M1: use \(u_4 = 164\) to set up a second equation in \(A\) and \(B\)

A1: cao

Notes

B1: cao

M1: correct form for the particular solution and substituted into recurrence relation

M1: compares coefficients and setting up all three equations in \(\lambda, \beta, \alpha\)

A1: correct general solution (or with consistent value of \(k\))

M1: use initial condition to obtain a quadratic equation in \(k\)

A1ft: correct solution for \(k\) following through their general solution

B1: correct explanation that the exponential term tends to zero as \(n\) becomes large

A1: cao

Alternative for third M mark: Note that candidates may calculate \(k\) immediately by eliminating \(u_1\) from \(2u_1 = u_0 - k\) and \(2u_2 = u_1 - 4k\) and comparing with \(4u_2 - u_0 = 27k^2\) to obtain a quadratic in \(k\)

Notes

1M1: attempt to write given information in terms of a recurrence equation (allow sign errors and errors in notation)

1A1*: CAO (NB equation is provided in the question so must see a correct unsimplified recurrence relation which is simplified with no errors to the required form for this mark).

Notes

1B1: CAO for auxiliary equation and corresponding solutions (this mark can be implied by correct complementary function)

2B1: complementary function CAO. Condone lack of ‘\(u_n =\)’.

1M1: substitutes \(u_n = \lambda n\) into their 2nd-order recurrence relation and solves to obtain \(\lambda = \ldots\)

Note: \(2\lambda n - \lambda(n - 1) - \lambda(n - 2) = 12\) o.e. \(\Rightarrow \lambda = \cdots\) can earn this mark

Note: Do not condone sign errors or errors in coefficients e.g.
\(2\lambda n - \lambda(n + 1) - \lambda(n - 2) = 12\) is M0
\(2\lambda n - 2\lambda(n - 1) - \lambda(n - 2) = 12)\) is M0

1A1: Correct general solution. Condone lack of ‘\(u_n =\)’.

2M1: Forms one equation in \(A\) and \(B\). General Solution must be of the form.
\(u_n = A + B\left(-\tfrac{1}{2}\right)^n + \mu n\) where \(\mu \neq 0\)

3dM1: Forms a second equation in \(A\) and \(B\). Dependent on the previous method mark.

2A1: Particular solution CAO. Do not condone lack of ‘\(u_n = \ldots\)’

Notes

M1: Obtains correct limit (ft their particular solution which must be of the correct form i.e. \(u_n = A + B\left(-\tfrac{1}{2}\right)^n + \mu n\) where \(\mu \neq 0\))

A1ft: Provides correct reasoning including comments relating to both of the following:

• \(n \to \infty, \left(-\tfrac{1}{2}\right)^n \to 0\)
• ‘\(-\tfrac{2}{3}\)’ becomes negligible (compared to \(4n\) as \(n \to \infty\))

Note: Condone ‘\(-\tfrac{2}{3}\) becomes insignificant’

Notes

B1: CAO for complementary function

M1: substituting correct trial solution into the recurrence relation – allow substitution of \(u_n = k\left(2^n\right)\) into \(u_n = 3u_{n-1} + 2^{n-1}\) but not \(u_n = 3u_{n-1} + 2^n\)

A1: CAO \(k = -1\)

A1: CAO for the general solution – must include \(u_n = \ldots\)

Notes

M1: using the condition \(u_1 = u_2\) to calculate a value for their \(A\) \(\left(= \frac{1}{3}\right)\) - this mark is dependent on the general solution being of the form \(\pm\lambda(3)^n \pm \mu(2)^n\)

A1: CAO for the particular solution (oe) – must include \(u_n = \ldots\) - however, if neither (general nor particular) solution is given in terms of \(u_n\) then award this mark if correct expression in terms of \(n\) seen (or if both solutions are given in terms of say \(u_{n+1}\))

Notes

B1: cao

B1: cao

Notes

B1: cao

M1: substituting their trial solution into the recurrence relation in an attempt to find their \(\lambda\)

A1: cao for the general solution

M1: using the conditions in the model to calculate \(A\)

A1: cao for the particular solution

Alternative approach for (b)

B1: \((1.02)^n u_0\)

M1: attempt sum of GP with \(a = 50\) and \(r = 1.02\) \(\left(u_n = \ldots + \dfrac{50\left(1 - 1.02^n\right)}{1 - 1.02}\right)\)

A1: general solution is \(u_n = (1.02)^n(u_0 + 2500) - 2500\) (or equivalent)

M1: Uses \(u_1 = 560\) to find \(u_0\) (e.g. \(560 = 1.02u_0 + 50 \Rightarrow u_0 = \ldots\))

A1: \(u_n = 3000(1.02)^n - 2500\)

Notes

M1: sets their particular solution greater than 3000 (condone equals) – their particular solution must be of the correct form \(\left(u_n = c(1.02)^n \pm d\right)\)

M1: dependent on previous M mark – re-arranging and correctly applies the process of taking logs for their particular solution

A1: cao (allow correct answer to 3 significant figures or 31)