A2 June 2021 Paper 2 Q4
4
(a) Show that\[(r + 1)^2 - r^2 = 2r + 1\] [1 mark]
(b) Use the method of differences to show that\[\sum_{r=1}^{n}(2r + 1) = n^2 + 2n\] [3 marks]
(c) Verify that using the formula for \(\displaystyle\sum_{r=1}^{n} r\) gives the same result as that given in part (b). [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Completes a rigorous argument to show the required result Must begin with \((r + 1)^2 - r^2 = \ldots\) | R1 | 2.1 |
| (1) |
Typical solution
\[\begin{aligned} (r + 1)^2 - r^2 &= r^2 + 2r + 1 - r^2 \\ &= 2r + 1 \text{ as required} \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Uses method of differences including at least the first two or last two terms | M1 | 1.1a |
| Identifies and simplifies the two remaining terms | M1 | 1.1a |
| Completes a rigorous argument to show the required result, including seeing at least the first two and the last two terms. Must begin with\[\sum_{r=1}^{n}(2r + 1) = \cdots\] | R1 | 2.1 |
| (3) |
Typical solution
\[\begin{aligned} \sum_{r=1}^{n}(2r + 1) &= \sum_{r=1}^{n}\left((r + 1)^2 - r^2\right) \\ &= \cancel{2^2} - 1^2 \\ &\quad + \cancel{3^2} - \cancel{2^2} \\ &\quad + \cdots \\ &\quad + \cdots \\ &\quad + \cancel{n^2} - \cancel{(n - 1)^2} \\ &\quad + (n + 1)^2 - \cancel{n^2} \end{aligned}\]\[\begin{aligned} &= (n + 1)^2 - 1 \\ &= n^2 + 2n + 1 - 1 \\ &= n^2 + 2n \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Recalls and states\[\sum_{r=1}^{n} r = \frac{1}{2}n(n + 1)\] | B1 | 1.2 |
| Splits the sum into two parts and uses their formula. | M1 | 1.1a |
| Completes a clear argument to show the required result. Condone the lack of limits on the summation signs. Must begin with \(\sum_{r=1}^{n}(2r + 1) = \cdots\) | R1 | 2.1 |
| (3) | ||
| (7 marks) |