A2 June 2024 Paper 1 Q6
6 The sequence \(u_1, u_2, u_3, \ldots\) is defined by
\[u_1 = 1\]\[u_{n+1} = u_n + 3n\]Prove by induction that for all integers \(n \geqslant 1\)
\[u_n = \frac{3}{2}n^2 - \frac{3}{2}n + 1\][4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Shows that \(u_1 = \dfrac{3}{2} \times 1^2 - \dfrac{3}{2} \times 1 + 1 = 1\) | B1 | 1.1b |
| Assumes the result is true for \(n = k\) and states \(u_{k+1} = \dfrac{3}{2}k^2 - \dfrac{3}{2}k + 1 + 3k\) at any point | M1 | 3.1a |
| Completes correct working to deduce that \(\dfrac{3}{2}k^2 - \dfrac{3}{2}k + 1 + 3k\) and \(\dfrac{3}{2}(k + 1)^2 - \dfrac{3}{2}(k + 1) + 1\) are equivalent | B1 | 2.2a |
| Concludes a reasoned argument by stating: \(u_n = \dfrac{3}{2}n^2 - \dfrac{3}{2}n + 1\) true for \(n = 1\) (seen anywhere) If \(u_n = \dfrac{3}{2}n^2 - \dfrac{3}{2}n + 1\) is true for \(n = k\), then true for \(n = k + 1\) Hence (by induction) \(u_n = \dfrac{3}{2}n^2 - \dfrac{3}{2}n + 1\) is true for all integers \(n \geqslant 1\) Condone “the result/formula” instead of \(u_n = \dfrac{3}{2}n^2 - \dfrac{3}{2}n + 1\) | R1 | 2.1 |
| (4 marks) |
Typical solution
Let \(n = 1\); then the formula gives
\[u_1 = \frac{3}{2} \times 1^2 - \frac{3}{2} \times 1 + 1 = 1\](So the formula is true for \(n = 1\))
Assume the formula is true for \(n = k\)
Then we want to show that
\[\begin{aligned}u_{k+1} &= \frac{3}{2}(k + 1)^2 - \frac{3}{2}(k + 1) + 1 \\ &= \frac{3}{2}(k^2 + 2k + 1 - k - 1) + 1 \\ &= \frac{3}{2}(k^2 + k) + 1\end{aligned}\]But
\[\begin{aligned}u_{k+1} &= \frac{3}{2}k^2 - \frac{3}{2}k + 1 + 3k \\ &= \frac{3}{2}(k^2 + k) + 1\end{aligned}\]as required
So the formula is also true for \(n = k + 1\)
The formula for \(u_n\) is true for \(n = 1\)
If the formula is true for \(n = k\), then the formula is also true for \(n = k + 1\)
Hence by induction \(u_n = \dfrac{3}{2}n^2 - \dfrac{3}{2}n + 1\) is true for all integers \(n \geqslant 1\)