## Question 1:

### Part 1(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{r(r+1)(r-1)} = \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1}$ | M1A1A1 (3) | M1: Attempt PFs by any valid method (by implication if 3 correct fractions seen). A1: any 2 fractions correct; A1: third fraction correct |

**Special Case:** $\frac{2}{r(r^2-1)} = \frac{2r}{r^2-1} - \frac{2}{r}$ seen, award M1A1A0. Award M1A0A0 provided of the form $\frac{2}{r(r^2-1)} = \frac{Ar}{r^2-1} - \frac{B}{r}$

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### Part 1(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $r=2$: $1 - \frac{2}{2} + \frac{1}{3}$; $r=3$: $\frac{1}{2} - \frac{2}{3} + \frac{1}{4}$; $r=4$: $\frac{1}{3} - \frac{2}{4} + \frac{1}{5}$ | M1 | Method of differences with at least 3 terms at start and 2 at end OR 2 at start and 3 at end. Must start at 2 and end at $n$ |
| $r=n-1$: $\frac{1}{n-2} - \frac{2}{n-1} + \frac{1}{n}$; $r=n$: $\frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1}$ | M1 | Last lines may be missing $k/(n-1)$ and $c/(n-2)$. These 2 M marks may be implied by correct extraction of terms. If starting from 1, M0M1 can be awarded |
| $\sum_{r=2}^{n}\left(\frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1}\right) = \left(1 - \frac{2}{2} + \frac{1}{2} + \frac{1}{n} - \frac{2}{n} + \frac{1}{n+1}\right)$ | A1 | Extract the remaining terms. $1 - 2/2$ may be missing and $1/n - 2/n$ may be combined |
| $\frac{1}{2}\sum_{r=1}^{n} \frac{2}{r(r+1)(r-1)} = \frac{1}{2} \times \left(\frac{1}{2} - \frac{1}{n} + \frac{1}{n+1}\right) = \frac{n^2+n-2}{4n(n+1)}$ | dM1A1 (5) | Include the $\frac{1}{2}$ and attempt a common denominator of the required form. Depends on both previous M marks. A1: $\frac{n^2+n-2}{4n(n+1)}$ |

**Special Case 1(b):** Terms listed as described above — award M1M1. Further progress unlikely as too many terms needed to establish the cancellation.

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