Standard +0.8 This is a two-part induction question from Further Maths FP1. Part (a) involves a summation with exponentials requiring algebraic manipulation of powers of 2, while part (b) requires proving a closed form for a recurrence relation involving two previous terms. Both parts demand careful algebraic handling and are more sophisticated than standard C1/C2 induction proofs, but follow established FP1 patterns without requiring novel insight.
9. (a) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\),
$$\sum _ { r = 1 } ^ { n } ( r + 1 ) 2 ^ { r - 1 } = n 2 ^ { n }$$
(b) A sequence of numbers is defined by
$$\begin{gathered}
u _ { 1 } = 0 , \quad u _ { 2 } = 32 , \\
u _ { n + 2 } = 6 u _ { n + 1 } - 8 u _ { n } \quad n \geqslant 1
\end{gathered}$$
Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\),
$$u _ { n } = 4 ^ { n + 1 } - 2 ^ { n + 3 }$$
If the result is true for \(n = k\) then it has been shown true for \(n = k+1\). As it is true for \(n = 1\) then it is true for all \(n\) (positive integers.)
A1
cso; statements can be seen anywhere in the solution. Do not award final A if \(n\) defined incorrectly e.g. '\(n\) is an integer' — award A0
Attempt \(u_{k+2}\) in terms of \(4^{k+2}\) and \(2^{k+5}\)
So \(u_{k+2} = 4^{(k+2)+1} - 2^{(k+2)+3}\)
A1
Correct expression
If the result is true for \(n = k\) and \(n = k+1\) then it has been shown true for \(n = k+2\). As it is true for \(n = 1\) and \(n = 2\) then it is true for all \(n\) (positive integers.)
A1
cso; statements can be seen anywhere in the solution. Do not award final A if \(n\) defined incorrectly e.g. '\(n\) is an integer' — award A0
(7 marks; Total: 12)
## Question 9:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $n=1$: rhs $=$ lhs $= 2$ | B1 | |
| Assume true for $n = k$ so $\sum_{r=1}^{k}(r+1)2^{r-1} = k2^k$; then $\sum_{r=1}^{k+1}(r+1)2^{r-1} = k2^k + (k+1+1)2^{k+1-1}$ | M1A1 | M1: Attempt to add $(k+1)^{\text{th}}$ term; A1: Correct expression |
| $= k2^k + (k+2)2^k = 2\times k2^k + 2\times 2^k \cdot\frac{1}{...}$... $= (k+1)2^{k+1}$ | A1 | At least one correct intermediate step required |
| If the result is **true for** $n = k$ then it has been shown **true for** $n = k+1$. As it is **true for** $n = 1$ then it is **true for all** $n$ (positive integers.) | A1 | cso; statements can be seen anywhere in the solution. Do not award final A if $n$ defined incorrectly e.g. '$n$ is an integer' — award A0 |
**(5 marks)**
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| When $n=1$: $u_1 = 4^2 - 2^4 = 0$ | B1 | $4^2 - 2^4 = 0$ seen |
| When $n=2$: $u_2 = 4^3 - 2^5 = 32$ | B1 | $4^3 - 2^5 = 32$ seen |
| Assume $u_k = 4^{k+1} - 2^{k+3}$ and $u_{k+1} = 4^{k+2} - 2^{k+4}$; $u_{k+2} = 6u_{k+1} - 8u_k$ | M1A1 | M1: Attempts $u_{k+2}$ in terms of $u_{k+1}$ and $u_k$; A1: Correct expression |
| $= 6(4^{k+2} - 2^{k+4}) - 8(4^{k+1} - 2^{k+3})$ $= 6\cdot4^{k+2} - 6\cdot2^{k+4} - 8\cdot4^{k+1} + 8\cdot2^{k+3}$ $= 6\cdot4^{k+2} - 3\cdot2^{k+5} - 2\cdot4^{k+2} + 2\cdot2^{k+5}$ $= 4\cdot4^{k+2} - 2^{k+5} = 4^{k+3} - 2^{k+5}$ | M1 | Attempt $u_{k+2}$ in terms of $4^{k+2}$ and $2^{k+5}$ |
| So $u_{k+2} = 4^{(k+2)+1} - 2^{(k+2)+3}$ | A1 | Correct expression |
| If the result is **true for** $n = k$ and $n = k+1$ then it has been shown **true for** $n = k+2$. As it is **true for** $n = 1$ and $n = 2$ then it is **true for all** $n$ (positive integers.) | A1 | cso; statements can be seen anywhere in the solution. Do not award final A if $n$ defined incorrectly e.g. '$n$ is an integer' — award A0 |
**(7 marks; Total: 12)**
9. (a) Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,
$$\sum _ { r = 1 } ^ { n } ( r + 1 ) 2 ^ { r - 1 } = n 2 ^ { n }$$
(b) A sequence of numbers is defined by
$$\begin{gathered}
u _ { 1 } = 0 , \quad u _ { 2 } = 32 , \\
u _ { n + 2 } = 6 u _ { n + 1 } - 8 u _ { n } \quad n \geqslant 1
\end{gathered}$$
Prove by induction that, for $n \in \mathbb { Z } ^ { + }$,
$$u _ { n } = 4 ^ { n + 1 } - 2 ^ { n + 3 }$$
\hfill \mbox{\textit{Edexcel FP1 2014 Q9 [12]}}