A2 October 2021 Paper 1 Q7
7 Prove that \(\displaystyle\sum_{r=1}^{n} \frac{r}{2^{r-1}} = 4 - \frac{n + 2}{2^{n-1}}\) for all \(n \geqslant 1\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{1} \frac{r}{2^{r-1}} = 1 = 4 - \frac{1 + 2}{2^0}\) so true for \(n = 1\) | B1 | 2.1 |
| Assume true for \(n = k\) so \(\displaystyle\sum_{r=1}^{k} \frac{r}{2^{r-1}} = 4 - \frac{k + 2}{2^{k-1}}\) | M1 | 2.1 |
| \(\displaystyle\sum_{r=1}^{k+1} \frac{r}{2^{r-1}} = 4 - \frac{k + 2}{2^{k-1}} + \frac{k + 1}{2^k}\) | M1 | 2.1 |
| \(= 4 - \dfrac{2k + 4 - k - 1}{2^k}\) | M1 | 1.1 |
| \(= 4 - \dfrac{k + 1 + 2}{2^k}\) so true for \(n = k + 1\) | A1 | 2.2a |
| So true for \(n = 1\) and if true for \(n = k\) then true for \(n = k + 1\) \(\Rightarrow\) true for all \(n\) | A1 | 2.2a |
| [6] |
Notes
A1: Dependent on all previous marks awarded