| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| 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 requiring routine base case, assumption, and inductive step. Parts (b)-(d) involve straightforward algebraic manipulation and substitution into the given formula, with minimal problem-solving insight needed. This is easier than average for Further Maths, as it's a multi-part scaffolded question testing basic proof technique and formula application. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| When \(n=1\): \(\sum_{r=1}^{1}r^2 = 1^2 = 1\) and \(\frac{1}{6}\times1\times2\times3 = 1\); \(\therefore\) true for \(n=1\) | B1 | Substitutes \(n=1\) into LHS and RHS of \(\sum_{r=1}^{n}r^2 = \frac{1}{6}n(n+1)(2n+1)\) |
| Assume true for \(n=k\): \(\sum_{r=1}^{k}r^2 = \frac{1}{6}k(k+1)(2k+1)\); considers \(\sum_{r=1}^{k}r^2 + (k+1)^2\) | M1 | Uses \(\sum_{r=1}^{k}r^2 = \frac{1}{6}k(k+1)(2k+1)\) and considers adding \((k+1)^2\); condone use of \(n\) in place of \(k\) |
| \(\Rightarrow \sum_{r=1}^{k+1}r^2 = \frac{1}{6}(k+1)(k(2k+1)+6(k+1)) = \frac{1}{6}(k+1)(2k^2+7k+6) = \frac{1}{6}(k+1)(k+2)(2k+3) = \frac{1}{6}(k+1)(k+1+1)(2(k+1)+1)\) | A1 | Completes working to show \(\frac{1}{6}k(k+1)(2k+1)+(k+1)^2\) is equivalent to \(\frac{1}{6}(k+1)(k+2)(2k+3)\); accept \(\frac{1}{6}(k+1)(k+2)(2k+3)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States true for \(n=1\), and if true for \(n=k\) then true for \(n=k+1\), and concludes by induction that \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) is true for all integers \(n \geq 1\) | 2.1 R1 | Dependent on all previous marks. Algebra must use an alternative letter to \(n\). Condone reference to 'rule'/'statement'/'it' in concluding statement. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{6} \times 2n(2n+1)(4n+1) = \frac{1}{3}n(2n+1)(4n+1)\) | 1.1b B1 | Accept partial factorisation e.g. \(\frac{n}{3}(8n^2+6n+1)\). ISW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{n}(2r)^2 = 4\sum_{r=1}^{n} r^2\) | 3.1a M1 | Writes required sum as a multiple of \(\sum_{r=1}^{n} r^2\) |
| \(= 4 \times \frac{1}{6}n(n+1)(2n+1) = \frac{2}{3}n(n+1)(2n+1)\) | 1.1b A1 | Accept partial factorisation e.g. \(\frac{2n}{3}(2n^2+3n+1)\). ISW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1^2+3^2+5^2+\cdots+(2n-1)^2 = \sum_{1}^{2n} r^2 - \sum_{1}^{n}(2r)^2\) | 3.1a M1 | Subtracts (c) from (b). Accept (c)\(-\)(b). Or writes sum of odd squares in terms of \(\sum r^2\) and \(\sum r\). |
| \(= \frac{1}{3}n(2n+1)(4n+1) - \frac{2}{3}n(n+1)(2n+1)\) \(= \frac{1}{3}n(2n+1)((4n+1)-2(n+1))\) \(= \frac{1}{3}n(2n-1)(2n+1)\) | 1.1b A1 | Correct expression in terms of \(n\) in any form. |
| Completes reasoned argument to obtain correct expression. | 2.1 R1 |
## Question 13:
**Part 13(a):**
| When $n=1$: $\sum_{r=1}^{1}r^2 = 1^2 = 1$ and $\frac{1}{6}\times1\times2\times3 = 1$; $\therefore$ true for $n=1$ | B1 | Substitutes $n=1$ into LHS and RHS of $\sum_{r=1}^{n}r^2 = \frac{1}{6}n(n+1)(2n+1)$ |
| Assume true for $n=k$: $\sum_{r=1}^{k}r^2 = \frac{1}{6}k(k+1)(2k+1)$; considers $\sum_{r=1}^{k}r^2 + (k+1)^2$ | M1 | Uses $\sum_{r=1}^{k}r^2 = \frac{1}{6}k(k+1)(2k+1)$ and considers adding $(k+1)^2$; condone use of $n$ in place of $k$ |
| $\Rightarrow \sum_{r=1}^{k+1}r^2 = \frac{1}{6}(k+1)(k(2k+1)+6(k+1)) = \frac{1}{6}(k+1)(2k^2+7k+6) = \frac{1}{6}(k+1)(k+2)(2k+3) = \frac{1}{6}(k+1)(k+1+1)(2(k+1)+1)$ | A1 | Completes working to show $\frac{1}{6}k(k+1)(2k+1)+(k+1)^2$ is equivalent to $\frac{1}{6}(k+1)(k+2)(2k+3)$; accept $\frac{1}{6}(k+1)(k+2)(2k+3)$ |
## Question 13 (Induction conclusion):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States true for $n=1$, **and** if true for $n=k$ then true for $n=k+1$, **and** concludes by induction that $\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)$ is true for all integers $n \geq 1$ | 2.1 R1 | Dependent on all previous marks. Algebra must use an alternative letter to $n$. Condone reference to 'rule'/'statement'/'it' in concluding statement. |
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## Question 13(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{6} \times 2n(2n+1)(4n+1) = \frac{1}{3}n(2n+1)(4n+1)$ | 1.1b B1 | Accept partial factorisation e.g. $\frac{n}{3}(8n^2+6n+1)$. ISW. |
---
## Question 13(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}(2r)^2 = 4\sum_{r=1}^{n} r^2$ | 3.1a M1 | Writes required sum as a multiple of $\sum_{r=1}^{n} r^2$ |
| $= 4 \times \frac{1}{6}n(n+1)(2n+1) = \frac{2}{3}n(n+1)(2n+1)$ | 1.1b A1 | Accept partial factorisation e.g. $\frac{2n}{3}(2n^2+3n+1)$. ISW. |
---
## Question 13(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1^2+3^2+5^2+\cdots+(2n-1)^2 = \sum_{1}^{2n} r^2 - \sum_{1}^{n}(2r)^2$ | 3.1a M1 | Subtracts (c) from (b). Accept (c)$-$(b). Or writes sum of odd squares in terms of $\sum r^2$ and $\sum r$. |
| $= \frac{1}{3}n(2n+1)(4n+1) - \frac{2}{3}n(n+1)(2n+1)$ $= \frac{1}{3}n(2n+1)((4n+1)-2(n+1))$ $= \frac{1}{3}n(2n-1)(2n+1)$ | 1.1b A1 | Correct expression in terms of $n$ in any form. |
| Completes reasoned argument to obtain correct expression. | 2.1 R1 | |
---13
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that, for all integers $n \geq 1$,
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$
[4 marks]\\
13
\item Hence, or otherwise, write down a factorised expression for the sum of the first $2 n$ squares
$$1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }$$
13
\item Use the formula in part (a) to write down a factorised expression for the sum of the first $n$ even squares
$$2 ^ { 2 } + 4 ^ { 2 } + 6 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }$$
13
\item Hence, or otherwise, show that the sum of the first $n$ odd squares is
$$a n ( b n - 1 ) ( b n + 1 )$$
where $a$ and $b$ are rational numbers to be determined.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2023 Q13 [10]}}