# Question 6:

| Answer | Mark | Guidance |
|--------|------|----------|
| For $k=1$, $1 \times 2^{1-1}$ and $1+(1-1)2^1 = 1$, so true for $k=1$ | B1 | |
| Assume true for $n = k$ | E1 | Explicit statement: 'assume true for $n=k$'; ignore irrelevant work |
| Next term is $(k+1)2^{k+1-1} = (k+1)2^k$ | M1, A1 | Attempt to find $(k+1)$th term; Correct |
| Add to both sides: RHS $= 1+(k-1)2^k + (k+1)2^k$ | M1 | Add to both sides |
| $= 1 + 2^k(k-1+k+1) = 1 + 2^k \times 2k = 1 + 2^{k+1}k$ | A1 | Correct simplification of RHS |
| $= 1 + ((k+1)-1)2^{k+1}$ | E1 | Statement: 'if true for $k$, true for $k+1$'; only give if simplification valid |
| Since true for $k=1$, true for $k = 1,2,3,\ldots$ by induction | E1, **[8]** | Relating to $k=1$, accept 'by induction' |

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