AS June 2019 Paper 1 Q5
5 Prove by induction that, for all positive integers \(n\), \(\displaystyle\sum_{r=1}^{n}\frac{1}{3^r} = \frac{1}{2}\left(1 - \frac{1}{3^n}\right)\). [6]
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(\tfrac{1}{3} = \tfrac{1}{2}\left(1 - \tfrac{1}{3}\right)\) so true for \(n = 1\) | B1 | 2.1 |
| Assume true for \(n = k\) so \(\displaystyle\sum_{r=1}^{k}\frac{1}{3^r} = \frac{1}{2}\left(1 - \frac{1}{3^k}\right)\) | M1 | 2.1 |
| so \(\displaystyle\sum_{r=1}^{k+1}\frac{1}{3^r} = \frac{1}{2}\left(1 - \frac{1}{3^k}\right) + \frac{1}{3^{k+1}}\) | M1 | 2.1 |
| \(= \dfrac{1}{2}\left(1 - \dfrac{3}{3^{k+1}} + \dfrac{2}{3^{k+1}}\right)\) | A1 | 1.1 |
| \(= \dfrac{1}{2}\left(1 - \dfrac{1}{3^{k+1}}\right)\) [so true for \(n = k + 1\)] | A1* | 2.2a |
| True for \(n = k \Rightarrow\) true for \(n = k + 1 \Rightarrow\) true for all \(n\) | E1dep | 2.4 |
| [6] |
Notes
M1: (1st) condone notation errors
M1: (2nd) adding \(\dfrac{1}{3^{k+1}}\) to \(\dfrac{1}{2}\left(1 - \dfrac{1}{3^k}\right)\)
A1: combining fractions
E1dep: dep A1*