# Question 9:

## Part (a)

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $n=1$, rhs $=$ lhs $= 2$ | B1 | |
| Assume true for $n = k$ so $\sum_{r=1}^{k}(r+1)2^{r-1} = k2^k$ | | |
| $\sum_{r=1}^{k+1}(r+1)2^{r-1} = k2^k + (k+1+1)2^{k+1-1}$ | M1A1 | M1: Attempt to add $(k+1)$th term. A1: Correct expression |
| $= k2^k + (k+2)2^k = 2 \times k2^k + 2 \times 2^k$ | | |
| $= (k+1)2^{k+1}$ | 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 |

## Part (b)

| Answer | Marks | 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 $u_{k+1} = 4^{k+2} - 2^{k+4}$ | | |
| $u_{k+2} = 6u_{k+1} - 8u_k = 6(4^{k+2} - 2^{k+4}) - 8(4^{k+1} - 2^{k+3})$ | M1A1 | M1: Attempts $u_{k+2}$ in terms of $u_{k+1}$ and $u_k$. A1: Correct expression |
| $= 6\cdot4^{k+2} - 6\cdot2^{k+4} - 8\cdot4^{k+1} + 8\cdot2^{k+3}$ | | |
| $= 6\cdot4^{k+2} - 3\cdot2^{k+5} - 2\cdot4^{k+2} + 2\cdot2^{k+5}$ | M1 | Attempt $u_{k+2}$ in terms of $4^{k+2}$ and $2^{k+5}$ |
| $= 4\cdot4^{k+2} - 2^{k+5} = 4^{k+3} - 2^{k+5}$ | A1 | Correct expression |
| So $u_{k+2} = 4^{(k+2)+1} - 2^{(k+2)+3}$ | | |
| 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 |