Standard +0.3 This is a standard proof by induction with a straightforward algebraic manipulation. The telescoping nature of 1/(r(r+1)) = 1/r - 1/(r+1) makes the inductive step mechanical, requiring only basic fraction algebra. While it's a Further Maths topic, it's one of the most routine induction proofs, slightly easier than average A-level questions overall.
When \(n=1\), LHS \(= \frac{1}{1\times2} = \frac{1}{2}\), RHS \(= \frac{1}{1+1} = \frac{1}{2}\). So LHS = RHS and result true for \(n=1\)
B1
Evaluate both sides for B1
Assume true for \(n=k\): \(\sum_{r=1}^{k}\frac{1}{r(r+1)} = \frac{k}{k+1}\) and so \(\sum_{r=1}^{k+1}\frac{1}{r(r+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}\)
Attempt common denominator for second M1 (dependent on first); \(\frac{k+1}{k+2}\) for A1
Result true for \(n=k+1\), and by induction true for \(n\in\mathbf{Z}^+\)
B1
'Assume true for \(n=k\)' and 'result true for \(n=k+1\)' and correct solution for final B1
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $n=1$, LHS $= \frac{1}{1\times2} = \frac{1}{2}$, RHS $= \frac{1}{1+1} = \frac{1}{2}$. So LHS = RHS and result true for $n=1$ | B1 | Evaluate both sides for B1 |
| Assume true for $n=k$: $\sum_{r=1}^{k}\frac{1}{r(r+1)} = \frac{k}{k+1}$ and so $\sum_{r=1}^{k+1}\frac{1}{r(r+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$ | M1 | Final two terms on second line for M1 |
| $\sum_{r=1}^{k+1}\frac{1}{r(r+1)} = \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$ | M1 A1 | Attempt common denominator for second M1 (dependent on first); $\frac{k+1}{k+2}$ for A1 |
| Result true for $n=k+1$, and by induction true for $n\in\mathbf{Z}^+$ | B1 | 'Assume true for $n=k$' and 'result true for $n=k+1$' and correct solution for final B1 |
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