| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation with exponentials |
| Difficulty | Challenging +1.2 Part (a) is a standard induction proof with a summation involving exponentials—requires careful algebraic manipulation in the inductive step but follows a routine template. Part (b) requires recognizing how to use part (a) to evaluate the sum from r=n+1 to 2n, involving substitution and algebraic simplification. While this is Further Maths content (inherently harder), it's a fairly standard FP2 induction question without requiring novel insight beyond applying the standard technique. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Assume result true for \(n=k\): \(\sum_{r=1}^{k}(r+1)2^{r-1} = k2^k\) | ||
| \(\sum_{r=1}^{k+1}(r+1)2^{r-1} = k2^k + (k+2)2^k\) | M1A1 | |
| \(= 2^k(k+k+2)\) | m1 | |
| \(= 2^k(2k+2)\) | ||
| \(= 2^{k+1}(k+1)\) | A1 | |
| \(n=1\): \(2 \times 2^0 = 2 = 1 \times 2^1\) | B1 | |
| \(P_k \Rightarrow P_{k+1}\) and \(P_1\) is true | E1 | Provided previous 5 marks earned |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{2n}(r+1)2^{r-1} - \sum_{r=1}^{n}(r+1)2^{r-1}\) | M1 | Sensible attempt at difference between 2 series |
| \(= 2n \cdot 2^{2n} - n2^n\) | A1 | |
| \(= n(2^{n+1}-1)2^n\) | A1 | AG |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Assume result true for $n=k$: $\sum_{r=1}^{k}(r+1)2^{r-1} = k2^k$ | | |
| $\sum_{r=1}^{k+1}(r+1)2^{r-1} = k2^k + (k+2)2^k$ | M1A1 | |
| $= 2^k(k+k+2)$ | m1 | |
| $= 2^k(2k+2)$ | | |
| $= 2^{k+1}(k+1)$ | A1 | |
| $n=1$: $2 \times 2^0 = 2 = 1 \times 2^1$ | B1 | |
| $P_k \Rightarrow P_{k+1}$ and $P_1$ is true | E1 | Provided previous 5 marks earned |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{2n}(r+1)2^{r-1} - \sum_{r=1}^{n}(r+1)2^{r-1}$ | M1 | Sensible attempt at difference between 2 series |
| $= 2n \cdot 2^{2n} - n2^n$ | A1 | |
| $= n(2^{n+1}-1)2^n$ | A1 | AG |
---4
\begin{enumerate}[label=(\alph*)]
\item Prove by induction that
$$2 + ( 3 \times 2 ) + \left( 4 \times 2 ^ { 2 } \right) + \ldots + ( n + 1 ) 2 ^ { n - 1 } = n 2 ^ { n }$$
for all integers $n \geqslant 1$.
\item Show that
$$\sum _ { r = n + 1 } ^ { 2 n } ( r + 1 ) 2 ^ { r - 1 } = n 2 ^ { n } \left( 2 ^ { n + 1 } - 1 \right)$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2006 Q4 [9]}}