# Question 6:

## Part (i):

| Working | Mark | Guidance |
|---------|------|----------|
| If $n=1$, $\begin{pmatrix}1&0\\-1&5\end{pmatrix}^1 = \begin{pmatrix}1&0\\-\frac{1}{4}(5^1-1)&5^1\end{pmatrix}$ so true for $n=1$ | B1 | Checks $n=1$ on both sides and states true for $n=1$ |
| Assume result true for $n=k$; $\begin{pmatrix}1&0\\-1&5\end{pmatrix}^{k+1} = \begin{pmatrix}1&0\\-\frac{1}{4}(5^k-1)&5^k\end{pmatrix}\begin{pmatrix}1&0\\-1&5\end{pmatrix}$ | M1 | Assumes true for $n=k$ and indicates intention to multiply power $k$ by power 1 either way |
| $= \begin{pmatrix}1&0\\-\frac{1}{4}(5^k-1)-5^k&5\times5^k\end{pmatrix}$ | M1 A1 | M1: Multiplies matrices, condone one slip. A1: Correct unsimplified matrix |
| $= \begin{pmatrix}1&0\\-\frac{1}{4}5^k+\frac{1}{4}-5^k&5^{k+1}\end{pmatrix} = \begin{pmatrix}1&0\\-\frac{1}{4}(5^{k+1}-1)&5^{k+1}\end{pmatrix}$ | A1 | Intermediate step required |
| True for $n=k+1$ if true for $n=k$, (and true for $n=1$) so true by induction for all $n\in\mathbb{Z}^+$ | A1cso | Must include statements in bold |

## Part (ii):

| Working | Mark | Guidance |
|---------|------|----------|
| If $n=1$, $\sum_{r=1}^{1}(2r-1)^2=1$ and $\frac{1}{3}n(4n^2-1)=1$, so true for $n=1$ | B1 | Checks $n=1$ on both sides and states true |
| Assume true for $n=k$ so $\sum_{r=1}^{k+1}(2r-1)^2 = \frac{1}{3}k(4k^2-1)+(2(k+1)-1)^2$ | M1 | Assumes true for $n=k$ and adds $(k+1)$th term |
| $= \frac{1}{3}(2k+1)\{(2k^2-k)+(3(2k+1))\}$ | M1 A1 | M1: Factorises out linear factor. A1: Correct with one linear and one quadratic factor |
| $= \frac{1}{3}(2k+1)\{2k^2+5k+3\}$ or $\frac{1}{3}(k+1)(4k^2+8k+3)$ | dA1 | Need to see $\frac{1}{3}(k+1)(4(k+1)^2-1)$ dependent on previous A1 |
| $= \frac{1}{3}(k+1)(2k+1)(2k+3) = \frac{1}{3}(k+1)(4(k+1)^2-1)$ | | |
| True for $n=k+1$ if true for $n=k$, (and true for $n=1$) so true by induction for all $n\in\mathbb{Z}^+$ | A1cso | Complete induction statement including bold statements |

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