**(a)** $\sum_{r=1}^{n}(6r-3)^2 = \sum_{r=1}^{n}(36r^2 - 36r + 9)$ | M1 | $\sum(r\sigma^2 + \beta r) = \alpha\sum r^2 + \beta\sum r$ seen/used

$= 36\sum_{r=1}^{n}r^2 - 36\sum_{r=1}^{n}r + 9\sum_{r=1}^{n}1$ | B1 | $\sum 1 = n$ seen or used

$= 36\sum_{r=1}^{n}r^2 - 36\sum_{r=1}^{n}r + 9n$ | | 

$= 36\left\{\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}\right\} + 9n$ | m1 | Subst of correct expressions for $\sum r^2$ and $\sum r$

$= 6n(n+1)[2n+1-3] + 9n$ | | 

$= 3n[4(n+1)(n-1) + 3]$ | A1 | OE Correct $3n[....]$ convincingly obtained before printed answer, where $[....]$ would reduce to $4n^2-1$ or $12n^3-3n$ obtained convincingly

$= 3n(4n^2 - 1)$ | A1 | AG. $3n(4n^2-1)$ convincingly obtained

**Total for (a): 5 marks**

**(b)** $\sum_{r=1}^{2n}r^3 = \frac{(2n)^2}{4}(2n+1)^2$ | B1 | OE correct expression for $\sum r^3$ stated or used

$\sum_{r=1}^{2n}r^3 - \sum_{r=1}^{n}(6r-3)^2 =$ | | 

$\frac{n^2(2n+1)^2 - 3n(4n^2-1)}{1} = n(2n+1)[n(2n+1) - 3(2n-1)]$ | M1 | Either $n(2n+1)g(n)$, where $g(n)$ is a quadratic OR reaching an equivalent stage in the factorisation of the correct quartic $4n^4 - 8n^3 + n^2 + 3n$ in $n$

$= n(2n+1)(2n^2 - 5n + 3)$ | A1 | OE A correct expression in $n$ with product containing at least two linear factors with the quadratic factor simplified

$= n(2n+1)(2n-3)(n-1)$ | A1 | Correct product of four linear factors in $n$

**Total for (b): 4 marks**

**Overall Total: 9 marks**

**Additional Notes:**
- (a) $\sum 9 = 9$ seen or used at any stage will result in B0A0A0
- (b) For penultimate A1 OEs include eg $n(n-1)(4n^2 - 4n - 3)$; $n(2n-3)(2n^2 - n - 1)$

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