## Question 6:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3r-2)^2 = 9r^2 - 12r + 4$ | B1 | Correct expansion |
| $\sum_{r=1}^{n}(9r^2 - 12r + 4) = 9 \times \frac{1}{6}n(n+1)(2n+1) - 12 \times \frac{1}{2}n(n+1) + \ldots$ | M1 | Substitutes at least one standard formula into expanded expression |
| $= 9 \times \frac{1}{6}n(n+1)(2n+1) - 12 \times \frac{1}{2}n(n+1) + 4n$ | A1 | Fully correct expression |
| $= \frac{1}{2}n\big[3(n+1)(2n+1) - 12(n+1) + 8\big]$ | dM1 | Attempts to factorise $\frac{1}{2}n$; dependent on first M mark and on there being an $n$ in all terms; having used at least one standard formula correctly |
| $= \frac{1}{2}n\big[6n^2 - 3n - 1\big]$* | A1* | Obtains printed result with no errors seen |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=5}^{n}(3r-2)^2 = \frac{1}{2}n(6n^2-3n-1) - \frac{1}{2}(4)(6(4)^2 - 3\times4 - 1)$ | M1 | Uses result from (a) substituting $n=4$ and subtracts from result in (a) to find first sum in terms of $n$ |
| $\sum_{r=1}^{28} r\cos\!\left(\frac{r\pi}{2}\right) = 0 - 2 + 0 + 4 + 0 - 6 + 0 + 8 + 0 - 10 + 0 + 12 + \ldots$ | M1 | Identifies the periodic nature of the second sum by calculating terms; may be implied by a sum of 14 |
| $3n^3 - \frac{3}{2}n^2 - \frac{1}{2}n - 166 + 103\times14 = 3n^3$, $\Rightarrow 3n^2 + n - 2552 = 0$ | A1 | Uses both sums and the given result to form the correct 3-term quadratic |
| $\Rightarrow 3n^2 + n - 2552 = 0 \Rightarrow n = \ldots$ | M1 | Solves their three-term quadratic to obtain at least one value for $n$ |
| $n = 29$ | A1 | Obtains $n=29$ only, or obtains $n=29$ and $n = -\frac{88}{3}$ and rejects $-\frac{88}{3}$ |

---