| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Prove sum formula |
| Difficulty | Moderate -0.8 This is a straightforward guided proof question where part (a) is trivial algebra, part (b) applies the standard method of differences with the identity already provided, and part (c) is routine verification using a known formula. All steps are heavily scaffolded with no problem-solving or insight required beyond following the prescribed method. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\) as required | R1 | Must begin with \((r+1)^2 - r^2 = \ldots\); rigorous argument required |
| Answer | Marks | Guidance |
|---|---|---|
| \[\sum_{r=1}^{n}(2r+1) = \sum_{r=1}^{n}((r+1)^2 - r^2)\] | M1 | Uses method of differences including at least first two or last two terms |
| \[= 2^2 - 1^2 + 3^2 - 2^2 + \cdots + n^2 - (n-1)^2 + (n+1)^2 - n^2\] | M1 | Identifies and simplifies the two remaining terms |
| \[= (n+1)^2 - 1 = n^2 + 2n + 1 - 1 = n^2 + 2n\] | R1 | Must begin with \(\sum_{r=1}^{n}(2r+1) = \cdots\); must see at least first two and last two terms |
| Answer | Marks | Guidance |
|---|---|---|
| \[\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\] | B1 | Recalls and states formula |
| \[\sum_{r=1}^{n}(2r+1) = 2 \times \frac{1}{2}n(n+1) + n\] | M1 | Splits sum into two parts and uses formula |
| \[= n^2 + n + n = n^2 + 2n \text{ as required}\] | R1 | Must begin with \(\sum_{r=1}^{n}(2r+1) = \cdots\); condone lack of limits on summation signs |
## Question 4(a):
$(r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1$ as required | R1 | Must begin with $(r+1)^2 - r^2 = \ldots$; rigorous argument required
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## Question 4(b):
$$\sum_{r=1}^{n}(2r+1) = \sum_{r=1}^{n}((r+1)^2 - r^2)$$ | M1 | Uses method of differences including at least first two or last two terms
$$= 2^2 - 1^2 + 3^2 - 2^2 + \cdots + n^2 - (n-1)^2 + (n+1)^2 - n^2$$ | M1 | Identifies and simplifies the two remaining terms
$$= (n+1)^2 - 1 = n^2 + 2n + 1 - 1 = n^2 + 2n$$ | R1 | Must begin with $\sum_{r=1}^{n}(2r+1) = \cdots$; must see at least first two and last two terms
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## Question 4(c):
$$\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$$ | B1 | Recalls and states formula
$$\sum_{r=1}^{n}(2r+1) = 2 \times \frac{1}{2}n(n+1) + n$$ | M1 | Splits sum into two parts and uses formula
$$= n^2 + n + n = n^2 + 2n \text{ as required}$$ | R1 | Must begin with $\sum_{r=1}^{n}(2r+1) = \cdots$; condone lack of limits on summation signs
---4
\begin{enumerate}[label=(\alph*)]
\item Show that
$$( r + 1 ) ^ { 2 } - r ^ { 2 } = 2 r + 1$$
4
\item Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } ( 2 r + 1 ) = n ^ { 2 } + 2 n$$
4
\item Verify that using the formula for $\sum _ { r = 1 } ^ { n } r$ gives the same result as that given in part (b).\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2021 Q4 [7]}}