FP1 January 2009 Q4
4. Prove by induction that, for \(n \in \mathbb{Z}^+\), \[\sum_{r=1}^{n}\frac{1}{r(r + 1)} = \frac{n}{n + 1}\] (5)
| Scheme | Marks |
|---|---|
| When \(n = 1\), LHS \(= \dfrac{1}{1 \times 2} = \dfrac{1}{2}\), RHS \(= \dfrac{1}{1 + 1} = \dfrac{1}{2}\). So LHS = RHS and result true for \(n = 1\) | B1 |
| Assume true for \(n = k\); \(\displaystyle\sum_{r=1}^{k}\frac{1}{r(r + 1)} = \frac{k}{k + 1}\) and so \(\displaystyle\sum_{r=1}^{k+1}\frac{1}{r(r + 1)} = \frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)}\) | M1 |
| \(\displaystyle\sum_{r=1}^{k+1}\frac{1}{r(r + 1)} = \frac{k(k + 2) + 1}{(k + 1)(k + 2)} = \frac{k^2 + 2k + 1}{(k + 1)(k + 2)} = \frac{(k + 1)^2}{(k + 1)(k + 2)} = \frac{k + 1}{k + 2}\) | M1 A1 |
| and so result is true for \(n = k + 1\) (and by induction true for \(n \in \mathbf{Z}^+\)) | B1 |
| [5] |
Notes
Evaluate both sides for first B1
Final two terms on second line for first M1
Attempt to find common denominator for second M1.
Second M1 dependent upon first.
\(\dfrac{k + 1}{k + 2}\) for A1
‘Assume true for \(n = k\)’ and ‘so result true for \(n = k + 1\)’ and correct solution for final B1