| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation with exponentials |
| Difficulty | Standard +0.3 This is a straightforward proof by induction with a product formula. Part (a) provides the key algebraic step needed for the inductive step, making part (b) mechanical: verify base case n=2, assume true for n=k, then multiply both sides by the (k+1)th term and apply the given identity. The algebra is routine and the structure is standard textbook induction, slightly easier than average A-level questions due to the scaffolding in part (a). |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\left(1 - \frac{1}{(k+1)^2}\right)\times\frac{k+1}{2k} = \frac{(k+1)^2-1}{(k+1)^2}\times\frac{k+1}{2k}\) | M1 | |
| \(= \frac{k^2+2k}{(k+1)^2}\times\frac{k+1}{2k}\) | A1 | |
| \(= \frac{k+2}{2(k+1)}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Assume true for \(n=k\): \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{(k+1)^2}\right)\) | M1 | |
| \(= \frac{k+2}{2(k+1)}\) | A1 | |
| True for \(n=2\) shown: \(1 - \frac{1}{2^2} = \frac{3}{4}\) | B1 | |
| \(P_n \Rightarrow P_{n+1}\) and \(P_2\) true | E1 | only if the other 3 marks earned |
## Question 6:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\left(1 - \frac{1}{(k+1)^2}\right)\times\frac{k+1}{2k} = \frac{(k+1)^2-1}{(k+1)^2}\times\frac{k+1}{2k}$ | M1 | |
| $= \frac{k^2+2k}{(k+1)^2}\times\frac{k+1}{2k}$ | A1 | |
| $= \frac{k+2}{2(k+1)}$ | A1 | AG |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| Assume true for $n=k$: $\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{(k+1)^2}\right)$ | M1 | |
| $= \frac{k+2}{2(k+1)}$ | A1 | |
| True for $n=2$ shown: $1 - \frac{1}{2^2} = \frac{3}{4}$ | B1 | |
| $P_n \Rightarrow P_{n+1}$ and $P_2$ true | E1 | only if the other 3 marks earned |
---6
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\left( 1 - \frac { 1 } { ( k + 1 ) ^ { 2 } } \right) \times \frac { k + 1 } { 2 k } = \frac { k + 2 } { 2 ( k + 1 ) }$$
\item Prove by induction that for all integers $n \geqslant 2$
$$\left( 1 - \frac { 1 } { 2 ^ { 2 } } \right) \left( 1 - \frac { 1 } { 3 ^ { 2 } } \right) \left( 1 - \frac { 1 } { 4 ^ { 2 } } \right) \ldots \left( 1 - \frac { 1 } { n ^ { 2 } } \right) = \frac { n + 1 } { 2 n }$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2007 Q6 [7]}}