# Question 8:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n}[r(r-1)-1] = \sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r - n$ | M1 | Split into separate sums |
| $= \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) - n$ | M1 | Use of at least one standard result (ignore 3rd term) |
| | A1 | Correct |
| $= \frac{1}{6}n[(n+1)(2n+1) - 3(n+1) - 6]$ | M1 | Attempt to factorise. If more than two errors, M0 |
| $= \frac{1}{6}n[2n^2 - 8]$ | | |
| $= \frac{1}{3}n[n^2 - 4]$ | A1 | Correct with factor $\frac{1}{3}n$ oe |
| $= \frac{1}{3}n(n+2)(n-2)$ | | Answer given |
| **[5]** | | |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $n=1$: $\sum_{r=1}^{n}[r(r-1)-1] = (1\times 0)-1 = -1$ and $\frac{1}{3}n(n+2)(n-2) = \frac{1}{3}\times 1\times 3\times -1 = -1$. So true for $n=1$ | B1 | |
| Assume true for $n=k$: $\sum_{r=1}^{k}[r(r-1)-1] = \frac{1}{3}k(k+2)(k-2)$ | E1 | Or "if true for $n=k$, then…" |
| $\Rightarrow \sum_{r=1}^{k+1}[r(r-1)-1] = \frac{1}{3}k(k+2)(k-2) + (k+1)k - 1$ | M1* | Add $(k+1)$th term to both sides |
| $= \frac{1}{3}k^3 + k^2 - \frac{4}{3}k + k - 1$ | | |
| $= \frac{1}{3}(k^3 + 3k^2 - k - 3)$ | | |
| $= \frac{1}{3}(k+1)(k^2+2k-3)$ | M1dep* | Attempt to factorise a cubic with 4 terms |
| $= \frac{1}{3}(k+1)(k+3)(k-1)$ | A1 | Or $= \frac{1}{3}n(n+2)(n-2)$ where $n=k+1$; or target seen |
| $= \frac{1}{3}(k+1)((k+1)+2)((k+1)-2)$ | | |
| But this is the given result with $n=k+1$ replacing $n=k$. Therefore if the result is true for $n=k$, it is also true for $n=k+1$. | E1 | Depends on A1 and first E1 |
| Since it is true for $n=1$, it is true for all positive integers $n$. | E1 | Depends on B1 and second E1 |
| **[7]** | | |

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