| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove recurrence relation formula |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring proof by induction of a closed form for a recurrence relation, followed by geometric series convergence and logarithmic summation. While induction itself is standard, the recurrence relation manipulation and the final part requiring simplification of a sum of logarithms to match a specific form with surds demands careful algebraic work beyond typical A-level questions. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04j Sum to infinity: convergent geometric series |r|<11.06f Laws of logarithms: addition, subtraction, power rules4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P_n\): \(u_n = 4\left(\frac{5}{4}\right)^n + 3\); Let \(n=1\) then \(4\left(\frac{5}{4}\right) + 3 = 8 \Rightarrow P_1\) true | B1 | States proposition; Proves base case |
| Assume \(P_k\) is true for some \(k\) | B1 | States inductive hypothesis |
| \(u_{k+1} = \frac{1}{4}\left(5\left(4\left(\frac{5}{4}\right)^k + 3\right) - 3\right) =\) correct step | M1 | Proves inductive step |
| \(= 4\left(\frac{5}{4}\right)^{k+1} + 3\) | A1 | |
| So \(P_k \Rightarrow P_{k+1}\). Therefore, by induction, \(P_n\) is true for all positive integers | A1 | States conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((u_n - 3)x^n = 4x^n\left(\frac{5}{4}\right)^n = 4\left(\frac{5x}{4}\right)^n\) so \(r = \left(\frac{5x}{4}\right)\) | M1 | |
| Series convergent for \(-1 < \frac{5x}{4} < 1 \Rightarrow -\frac{4}{5} < x < \frac{4}{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{n=1}^{N}\ln(u_n - 3) = \sum_{n=1}^{N}\ln\left(4\left(\frac{5}{4}\right)^n\right) = \left(\ln\frac{5}{4}\right)\sum_{n=1}^{N}n + \sum_{n=1}^{N}\ln 4\) | M1 | Alt: \(\sum_{n=1}^{N}\ln(u_n-3) = \ln\prod 4\left(\frac{5}{4}\right)^n\) |
| \(= \frac{1}{2}N(N+1)\ln\frac{5}{4} + N\ln 4\), using \(\sum_{n=1}^{N}n = \frac{1}{2}N(N+1)\) | M1 | \(= N\ln 4 + \frac{N(N+1)}{2}\ln\left(\frac{5}{4}\right)\) |
| \(= N^2\ln\frac{\sqrt{5}}{2} + N\ln(2\sqrt{5}) \Rightarrow a = \frac{\sqrt{5}}{2},\ b = 2\sqrt{5}\) | A1 |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P_n$: $u_n = 4\left(\frac{5}{4}\right)^n + 3$; Let $n=1$ then $4\left(\frac{5}{4}\right) + 3 = 8 \Rightarrow P_1$ true | B1 | States proposition; Proves base case |
| Assume $P_k$ is true for some $k$ | B1 | States inductive hypothesis |
| $u_{k+1} = \frac{1}{4}\left(5\left(4\left(\frac{5}{4}\right)^k + 3\right) - 3\right) =$ correct step | M1 | Proves inductive step |
| $= 4\left(\frac{5}{4}\right)^{k+1} + 3$ | A1 | |
| So $P_k \Rightarrow P_{k+1}$. Therefore, by induction, $P_n$ is true for all positive integers | A1 | States conclusion |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(u_n - 3)x^n = 4x^n\left(\frac{5}{4}\right)^n = 4\left(\frac{5x}{4}\right)^n$ so $r = \left(\frac{5x}{4}\right)$ | M1 | |
| Series convergent for $-1 < \frac{5x}{4} < 1 \Rightarrow -\frac{4}{5} < x < \frac{4}{5}$ | A1 | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{n=1}^{N}\ln(u_n - 3) = \sum_{n=1}^{N}\ln\left(4\left(\frac{5}{4}\right)^n\right) = \left(\ln\frac{5}{4}\right)\sum_{n=1}^{N}n + \sum_{n=1}^{N}\ln 4$ | M1 | Alt: $\sum_{n=1}^{N}\ln(u_n-3) = \ln\prod 4\left(\frac{5}{4}\right)^n$ |
| $= \frac{1}{2}N(N+1)\ln\frac{5}{4} + N\ln 4$, using $\sum_{n=1}^{N}n = \frac{1}{2}N(N+1)$ | M1 | $= N\ln 4 + \frac{N(N+1)}{2}\ln\left(\frac{5}{4}\right)$ |
| $= N^2\ln\frac{\sqrt{5}}{2} + N\ln(2\sqrt{5}) \Rightarrow a = \frac{\sqrt{5}}{2},\ b = 2\sqrt{5}$ | A1 | |
---9 For the sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$, it is given that $u _ { 1 } = 8$ and
$$u _ { r + 1 } = \frac { 5 u _ { r } - 3 } { 4 }$$
for all $r$.\\
(i) Prove by mathematical induction that
$$u _ { n } = 4 \left( \frac { 5 } { 4 } \right) ^ { n } + 3$$
for all positive integers $n$.\\
(ii) Deduce the set of values of $x$ for which the infinite series
$$\left( u _ { 1 } - 3 \right) x + \left( u _ { 2 } - 3 \right) x ^ { 2 } + \ldots + \left( u _ { r } - 3 \right) x ^ { r } + \ldots$$
is convergent.\\
(iii) Use the result given in part (i) to find surds $a$ and $b$ such that
$$\sum _ { n = 1 } ^ { N } \ln \left( u _ { n } - 3 \right) = N ^ { 2 } \ln a + N \ln b .$$
\hfill \mbox{\textit{CAIE FP1 2018 Q9 [10]}}