| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Challenging +1.2 Part (a) is a routine application of standard sum formulae requiring algebraic manipulation to match a given form. Part (b) requires the insight to use difference of sums (sum from 1 to 5n minus sum from 1 to n), then factorisation. This is a standard Further Maths technique but requires more problem-solving than typical A-level questions. The 4+marks suggests moderate length but the techniques are well-practiced in Further Maths courses. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{n}r(r+3) = \sum_{r=1}^{n}r^2 + 3\sum_{r=1}^{n}r\) | M1 (AO1.1a) | Writes sum in form \(\alpha\sum r^2 + \beta\sum r\) where \(\alpha\) and \(\beta\) are numbers. PI by use of correct standard formulae. |
| \(= \frac{1}{6}n(n+1)(2n+1) + 3\times\frac{1}{2}n(n+1)\) | A1 (AO1.2) | Recalls and uses standard formulae to obtain correct expression in terms of \(n\). May be unsimplified. |
| \(= \frac{1}{6}n(n+1)(2n+1+9) = \frac{1}{6}n(n+1)(2n+10)\) | M1 (AO1.1a) | Identifies \(n\) and \((n+1)\) as common factors. Must be at least one correct term inside remaining bracket. |
| \(= \frac{1}{3}n(n+1)(n+5)\) | R1 (AO2.1) | Completes fully correct proof. Must have \(a=\frac{1}{3}\) and \(b=5\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=n+1}^{5n}r(r+3) = \sum_{r=1}^{5n}r(r+3) - \sum_{r=1}^{n}r(r+3)\) | M1 (AO1.1a) | Substitutes \(5n\) into expression of form \(an(n+1)(n+b)\) and subtracts part (a). |
| \(= \frac{1}{3}(5n)(5n+1)(5n+5) - \frac{1}{3}n(n+1)(n+5)\) \(= \frac{1}{3}n(n+1)(25(5n+1)-(n+5))\) | A1 (AO1.1b) | Obtains correct expression. May be unsimplified. |
| \(= \frac{1}{3}n(n+1)(124n+20) = \frac{4}{3}n(n+1)(31n+5)\) | A1 (AO1.1b) | Obtains correct fully factorised expression. |
## Question 9(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}r(r+3) = \sum_{r=1}^{n}r^2 + 3\sum_{r=1}^{n}r$ | M1 (AO1.1a) | Writes sum in form $\alpha\sum r^2 + \beta\sum r$ where $\alpha$ and $\beta$ are numbers. PI by use of correct standard formulae. |
| $= \frac{1}{6}n(n+1)(2n+1) + 3\times\frac{1}{2}n(n+1)$ | A1 (AO1.2) | Recalls and uses standard formulae to obtain correct expression in terms of $n$. May be unsimplified. |
| $= \frac{1}{6}n(n+1)(2n+1+9) = \frac{1}{6}n(n+1)(2n+10)$ | M1 (AO1.1a) | Identifies $n$ and $(n+1)$ as common factors. Must be at least one correct term inside remaining bracket. |
| $= \frac{1}{3}n(n+1)(n+5)$ | R1 (AO2.1) | Completes fully correct proof. Must have $a=\frac{1}{3}$ and $b=5$. |
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## Question 9(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=n+1}^{5n}r(r+3) = \sum_{r=1}^{5n}r(r+3) - \sum_{r=1}^{n}r(r+3)$ | M1 (AO1.1a) | Substitutes $5n$ into expression of form $an(n+1)(n+b)$ and subtracts part (a). |
| $= \frac{1}{3}(5n)(5n+1)(5n+5) - \frac{1}{3}n(n+1)(n+5)$ $= \frac{1}{3}n(n+1)(25(5n+1)-(n+5))$ | A1 (AO1.1b) | Obtains correct expression. May be unsimplified. |
| $= \frac{1}{3}n(n+1)(124n+20) = \frac{4}{3}n(n+1)(31n+5)$ | A1 (AO1.1b) | Obtains correct fully factorised expression. |9
\begin{enumerate}[label=(\alph*)]
\item Use the standard formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ to show that
$$\sum _ { r = 1 } ^ { n } r ( r + 3 ) = a n ( n + 1 ) ( n + b )$$
where $a$ and $b$ are constants to be determined.\\[0pt]
[4 marks]\\
9
\item Hence, or otherwise, find a fully factorised expression for
$$\sum _ { r = n + 1 } ^ { 5 n } r ( r + 3 )$$
$$\mathbf { A } = \left[ \begin{array} { c c }
3 & i - 1 \\
i & 2
\end{array} \right]$$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q9 [7]}}