A2 October 2021 Q6
6. A recurrence system is defined by
\[u_{n+2} = 9(n+1)^2 u_n - 3u_{n+1} \qquad n \geqslant 1\]\[u_1 = -3,\ u_2 = 18\]Prove by induction that, for \(n \in \mathbb{N}\),
\[u_n = (-3)^n n!\](6)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1 : u_1 = (-3)^1 \times 1! = -3\) \(n = 2 : u_2 = (-3)^2 \times 2! = 9 \times 2 = 18\) Hence true for \(n = 1\) and \(n = 2\) | B1 | 2.2a |
| Assume true for some \(n = k\) and \(n = k + 1\), so \(u_k = (-3)^k k!\) and \(u_{k+1} = (-3)^{k+1}(k+1)!\) | M1 | 2.4 |
| Then \(u_{k+2} = 9(k+1)^2\left((-3)^k k!\right) - 3\left((-3)^{k+1}(k+1)!\right)\) | M1 | 1.1b |
| \(= (-3)^k k!\left[9(k+1)^2 - 3(-3)(k+1)\right]\) | M1 | 1.1b |
| \(= (-3)^k k!\left[9(k+1)(k+1+1)\right] = (-3)^k \times (-3)^2 \times (k+1)(k+2)k!\) \(= (-3)^{k+2}(k+2)!\) | A1 | 2.1 |
| Hence if true for \(n = k\) and \(n = k + 1\) then true for \(n = k + 2\). As also true for \(n = 1\) and \(n = 2\), then true for all \(n \in \mathbb{N}\) by mathematical induction. | A1 | 2.4 |
| (6) | ||
| (6 marks) |
Notes
B1: Checks the closed form works for \(n = 1\) and \(n = 2\)
M1: Makes the inductive assumption. May use e.g. \(n = k - 2\) and \(n = k - 1\) instead and show true for \(n = k\). It must be clear it is the closed forms they are assuming, not a recurrence form.
M1: Substitutes expression for \(n = k\) and \(n = k + 1\) (or equivalents) into the recurrence formula.
M1: Takes out common factors of at least \((-3)^k k!\) in their expression, or equivalent for their assumed true values. Treatment of the \((-3)\) must be correct, but condone invisible brackets if recovered.
Note: they may well take out more at this stage, which is fine, e.g.
\(u_{k+2} = 9(k+1)^2\left((-3)^k k!\right) - 3\left((-3)^{k+1}(k+1)!\right) = (-3)^{k+2}(k+1)!\left[(k+1) + 1\right]\)
A1: Simplifies correctly to the required form for their assumed true values.
A1: Correct conclusion made. Depends on all three M’s and the A being gained. Must convey the ideas of 1) true for \(n = 1\) and \(n = 2\), 2) if true for two successive cases, it is also true for the next case and 3) a suitable conclusion that it is true for all positive \(n\).