**(a)** If $n = 1$, $\sum_{r=1}^{1} r^2 = 1$ and $\frac{1}{6}n(n+1)(2n+1) = \frac{1}{6} \times 1 \times 2 \times 3 = 1$, so true for $n = 1$. | **B1** | 

Assume result true for $n = k$ | **M1** | 

$\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2$ | **M1** | 

$= \frac{1}{6}(k+1)(2k^2 + 7k + 6)$ or $= \frac{1}{6}(k+2)(2k^2 + 5k + 3)$ or $= \frac{1}{6}(2k+3)(k^2 + 3k + 2)$ | **A1** | 

$= \frac{1}{6}(k+1)(k+2)(2k+3) = \frac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)$ or equivalent | **dM1** | 

True for $n = k + 1$ if true for $n = k$, (and true for $n = 1$) so true by induction for all $n$. | **A1 cso** | (6 marks)

**Alternative for (a)** After first three marks B M M1 as earlier:

May state RHS $= \frac{1}{6}(k+1)((k+1)+1)(2(k+1)+1) = \frac{1}{6}(k+1)(k+2)(2k+3)$ for third M1

Expands to $\frac{1}{6}(k+1)(2k^2 + 7k + 6)$ and show equal to $\sum_{r=1}^{k+1} r^2 = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2$ for A1

So true for $n = k+1$ if true for $n = k$, and true for $n = 1$, so by induction for all $n$. | **B1M1M1, dM1, A1, A1 cso** | (6 marks)

**(b)** $\sum_{r=1}^{n}(r^2 + 5r + 6) = \sum_{r=1}^{n} r^2 + 5\sum_{r=1}^{n} r + \left(\sum_{r=1}^{n} 6\right)$ | **M1** | 

$\frac{1}{6}n(n+1)(2n+1) + \frac{5}{2}n(n+1), + 6n$ | **A1, B1** | 

$= \frac{1}{6}n[(n+1)(2n+1) + 15(n+1) + 36]$ | **M1** | 

$= \frac{1}{6}n[2n^2 + 18n + 52] = \frac{1}{3}n(n^2 + 9n + 26)$ or $a = 9, b = 26$ | **A1** | (5 marks)

**(c)** $\sum_{r=n+1}^{2n}(r+2)(r+3) = \frac{1}{3} \cdot 2n(4n^2 + 18n + 26) - \frac{1}{3}n(n^2 + 9n + 26)$ | **M1 A1 ft** | 

$\frac{1}{3}n(8n^2 + 36n + 52 - n^2 - 9n - 26) = \frac{1}{3}n(7n^2 + 27n + 26)$ | **A1 cso** | (3 marks)

**Total: 14 marks**

**Notes:**

(a) B1: Checks $n = 1$ on both sides and states true for $n = 1$ here or in conclusion

M1: **Assumes true for $n = k$** (should use one of these two words)

M1: Adds $(k+1)$th term to sum of $k$ terms

A1: Correct work to support proof

M1: Deduces $\frac{1}{6}n(n+1)(2n+1)$ with $n = k+1$

A1: Makes induction statement. Statement $n = 1$ here could contribute to B1 mark earlier

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## Question 9 Notes (Continued)

(b) M1: Expands and splits (but allow $\Sigma 6$ rather than sigma $6$ for this mark)

A1: first two terms correct

B1: for $6n$

M1: Take out factor $n/6$ or $n/3$ correctly – no errors factorising

A1: for correct factorised cubic or for identifying $a$ and $b$

(c) M1: Try to use $\sum_{r=n+1}^{2n}(r+2)(r+3) = \sum_{r=1}^{2n}(r+2)(r+3) - \sum_{r=1}^{n}(r+2)(r+3)$ with previous result used **at least once**

A1 ft: Two correct expressions for their $a$ and $b$ values

A1: Completely correct work to printed answer