AS June 2021 Paper 1 Q13
13 Prove by induction that, for all integers \(n \geqslant 1\)
\[\sum_{r=1}^{n} 2^{-r} = 1 - 2^{-n}\][4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Demonstrates the rule is correct for \(n = 1\) | B1 | 2.2a |
| Assumes the rule is true for \(n = k\) and adds \(2^{-(k+1)}\) to \(1 - 2^{-k}\) Condone incorrect or missing brackets for this mark only. | M1 | 2.4 |
| Correctly obtains \(1 - 2^{-(k+1)}\) from \(1 - 2^{-k} + 2^{-(k+1)}\) | A1 | 2.2a |
| Concludes a reasoned argument by stating that The rule is true for \(n = 1\); that if the rule is true for \(n = k\) then it is also true for \(n = k + 1\) and hence, by induction, the rule is true for all integers \(n \geqslant 1\). | R1 | 2.1 |
| (4 marks) |
Typical solution
Try \(n = 1\):
\[\text{LHS} = 2^{-1} = \tfrac{1}{2}\]and
\[\text{RHS} = 1 - 2^{-1} = \tfrac{1}{2}\]LHS = RHS \(\therefore\) the rule is true for \(n = 1\)
Assume the rule is true for \(n = k\):
\[\sum_{r=1}^{k} 2^{-r} = 1 - 2^{-k}\]\[\sum_{r=1}^{k} 2^{-r} + 2^{-(k+1)} = 1 - 2^{-k} + 2^{-(k+1)}\]\[\sum_{r=1}^{k+1} 2^{-r} = 1 + 2^{-(k+1)}(-2^1 + 1)\]\[\sum_{r=1}^{k+1} 2^{-r} = 1 - 2^{-(k+1)}\]\(\therefore\) the rule is also true for \(n = k + 1\)
As the rule is true for \(n = 1\), and if true for \(n = k\) then it is also true for \(n = k + 1\), then by induction the rule is true for all integers \(n \geqslant 1\)