| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation formula |
| Difficulty | Standard +0.3 Part (a) is a standard textbook induction proof of a well-known formula. Part (b) uses a telescoping sum technique that, while requiring careful algebraic manipulation, follows a prescribed identity. Part (c) is a routine limit calculation. This is a typical Further Maths question requiring multiple techniques but no novel insight, making it slightly easier than average overall. |
| Spec | 4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1^2 = \frac{1}{6}(1)(2)(3)\) so \(H_1\) is true. | B1 | Checks base case. |
| Assume that \(\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)\). | B1 | States inductive hypothesis. |
| \(\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2\) | M1 | Considers sum to \(k+1\). |
| \(\frac{1}{6}(k+1)(2k^2+k+6k+6) = \frac{1}{6}(k+1)(2k^2+7k+6) = \frac{1}{6}(k+1)(k+2)(2k+3)\) | A1 | |
| So \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\). | A1 | States conclusion. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3^5 - 1^5 + 5^5 - 3^5 + 7^5 - 5^5 \ldots + (2n+1)^5 - (2n-1)^5\) | M1 | Applies method of differences to LHS, writes complete terms for at least three values of \(r\) including first and last. |
| \((2n+1)^5 - 1\) | A1 | SCB1 for insufficient complete terms shown. |
| \(160S_n + \frac{40}{3}n(n+1)(2n+1) + 2n\) | M1 A1 | Sums RHS and applies standard formulae. |
| \(160S_n = (2n+1)^5 - \frac{40}{3}n(n+1)(2n+1) - (2n+1)\) \(160S_n = (2n+1)\left((2n+1)^4 - \frac{40}{3}n(n+1) - 1\right)\) | M1 | Makes \(S_n\) the subject and factorises. |
| \((2n+1)\left(16n^4+32n^3+\frac{32}{3}n^2-\frac{16}{3}n\right) = \frac{16}{3}n(2n+1)(3n^3+6n^2+2n-1)\) \(\Rightarrow S_n = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n^{-5}S_n = \frac{1}{30}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\left(3+\frac{3}{n}-\frac{1}{n^2}\right)\) | M1 | Divides by \(n^5\). |
| \(\frac{1}{5}\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1^2 = \frac{1}{6}(1)(2)(3)$ so $H_1$ is true. | **B1** | Checks base case. |
| Assume that $\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k+1)(2k+1)$. | **B1** | States inductive hypothesis. |
| $\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2$ | **M1** | Considers sum to $k+1$. |
| $\frac{1}{6}(k+1)(2k^2+k+6k+6) = \frac{1}{6}(k+1)(2k^2+7k+6) = \frac{1}{6}(k+1)(k+2)(2k+3)$ | **A1** | |
| So $H_{k+1}$ is true. By induction, $H_n$ is true for all positive integers $n$. | **A1** | States conclusion. |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3^5 - 1^5 + 5^5 - 3^5 + 7^5 - 5^5 \ldots + (2n+1)^5 - (2n-1)^5$ | **M1** | Applies method of differences to LHS, writes complete terms for at least three values of $r$ including first and last. |
| $(2n+1)^5 - 1$ | **A1** | SCB1 for insufficient complete terms shown. |
| $160S_n + \frac{40}{3}n(n+1)(2n+1) + 2n$ | **M1 A1** | Sums RHS and applies standard formulae. |
| $160S_n = (2n+1)^5 - \frac{40}{3}n(n+1)(2n+1) - (2n+1)$ $160S_n = (2n+1)\left((2n+1)^4 - \frac{40}{3}n(n+1) - 1\right)$ | **M1** | Makes $S_n$ the subject and factorises. |
| $(2n+1)\left(16n^4+32n^3+\frac{32}{3}n^2-\frac{16}{3}n\right) = \frac{16}{3}n(2n+1)(3n^3+6n^2+2n-1)$ $\Rightarrow S_n = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$ | **A1** | AG |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n^{-5}S_n = \frac{1}{30}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\left(3+\frac{3}{n}-\frac{1}{n^2}\right)$ | **M1** | Divides by $n^5$. |
| $\frac{1}{5}$ | **A1** | |
---4
\begin{enumerate}[label=(\alph*)]
\item Prove by mathematical induction that, for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
\includegraphics[max width=\textwidth, alt={}, center]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-08_2716_35_143_2012}
The sum $S _ { n }$ is defined by $S _ { n } = \sum _ { r = 1 } ^ { n } r ^ { 4 }$.
\item Using the identity
$$( 2 r + 1 ) ^ { 5 } - ( 2 r - 1 ) ^ { 5 } \equiv 160 r ^ { 4 } + 80 r ^ { 2 } + 2$$
show that $S _ { n } = \frac { 1 } { 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ { 2 } + 3 n - 1 \right)$.
\item Find the value of $\lim _ { n \rightarrow \infty } \left( n ^ { - 5 } S _ { n } \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q4 [13]}}