AS October 2021 Paper 1 Q5
5 Prove by induction that \(\displaystyle\sum_{r=1}^{n} r \times 2^{r-1} = 1 + (n - 1)2^n\) for all positive integers \(n\). [5]
| Scheme | Marks | AO |
|---|---|---|
| Check \(n = 1\): \(1 \times 2^0 = 1 = 1 + 0 \times 2\) | B1 | 2.1 |
| Assume \(n = k\) or \(\displaystyle\sum_{r=1}^{k} r \times 2^{r-1} = 1 + (k - 1)2^k\) | B1 | 2.1 |
| \(\displaystyle\sum_{r=1}^{k+1} r \times 2^{r-1} = 1 + (k - 1)2^k + (k + 1)2^k\) | M1 | 2.1 |
| \(= 1 + 2k \times 2^k = 1 + k \times 2^{k+1}\) which is the result for \(n = k + 1\) | A1* | 2.2a |
| True for \(n = 1\), and if true for \(n = k\) then also true for \(n = k + 1\), so true for all \(n\) | B1dep | 2.2a |
| [5] |
Notes
B1: (2nd) soi from correct next step
but \(\displaystyle\sum_{r=1}^{k} k \times 2^{k-1}\) is B0
M1: condone missing brackets if intention is clear subsequently
B1dep: dep A1*