## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $n=1$, $(-1)^0 \frac{1\times 2}{2} = 1$ and $1^2 = 1$, so true for $n=1$ | B1 | |
| Assume true for $n=k$: $\Rightarrow 1^2 - 2^2 + 3^2 - \ldots + (-1)^{k-1}k^2 = (-1)^{k-1}\frac{k(k+1)}{2}$ | E1 | Assuming true result for some $n$. Condone series shown incomplete |
| $\Rightarrow 1^2 - 2^2 + 3^2 - \ldots + (-1)^{k-1}k^2 + (-1)^{k+1-1}(k+1)^2$ | M1* | Adding $(k+1)$th term to both sides |
| $= (-1)^{k-1}\frac{k(k+1)}{2} + (-1)^{k+1-1}(k+1)^2$ | | |
| $= (-1)^k\left[\frac{-k(k+1)}{2} + (k+1)^2\right]$ | M1 Dep* | Attempt to factorise (at least one valid factor) |
| $= (-1)^k(k+1)\left(\frac{-k}{2} + k + 1\right)$ | A1 | Correct factorisation. Accept $(-1)^{k\pm m}$ provided expression correct |
| $= (-1)^k(k+1)\left(\frac{k+2}{2}\right)$ | A1 | Valid simplification with $(-1)^k$ |
| $= (-1)^{[n-1]}\frac{n(n+1)}{2}$, $n = k+1$ | E1 | Or target seen. Dependent on A1 and previous E1 |
| Therefore if true for $n=k$ it is also true for $n=k+1$. Since it is true for $n=1$, it is true for all positive integers $n$. | E1 | Dependent on B1 and previous E1 |
| | **[8]** | |

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