## Question 14:

### Main Algebraic Method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n$ is even: $(2k+1)^3 - (2k)^3 = 8k^3 + 12k^2 + 6k + 1 - 8k^3 = 6k(2k+1)+1 \Rightarrow$ odd | M1 | Attempt $(n+1)^3 - n^3$ when $n=2k$ **or** $n=2k\pm1$; multiply out to three-term quadratic; condone arithmetical slips |
| When $n$ is odd: $(2k+2)^3-(2k+1)^3 = 6k(2k+3)+7 \Rightarrow$ odd | A1 | Complete argument for $n=2k$ **or** $n=2k+1$; correct simplified quadratic; reason why odd |
| Both cases shown | dM1 | Attempts both $n=2k$ **and** $n=2k\pm1$ |
| Hence odd for all $n(\in\mathbb{N})$ | A1* | Complete argument for **both** cases with overall conclusion; withhold if $n$ used instead of $k$, or $n\in\mathbb{R}$ (but $n\in\mathbb{Z}^+$ acceptable) |

### Alternative — Algebraic with logic:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Multiply out and simplify to three-term quadratic | M1 | Condone arithmetical slips |
| $3n^2 + 3n + 1$ | A1 | Correct expression |
| Factorise so that $n^2+n \to n(n+1)$, e.g. $3n(n+1)+1$ | dM1 | |
| $n(n+1)$ always even (product of consecutive integers), so $3n(n+1)$ is odd$\times$even $=$ even, thus $3n(n+1)+1$ is odd for all $n(\in\mathbb{N})$ | A1* | Full explanation required |

### Proof by Contradiction:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Multiply out to three-term quadratic | M1 | |
| $3n^2 + 3n + 1$ | A1 | |
| Set $3n^2+3n+1=2k \Rightarrow 3n(n+1)=2k-1$ | dM1 | |
| $n(n+1)$ always even so $3n(n+1)$ is even, but $2k-1$ is odd — contradiction; hence $(n+1)^3-n^3$ is odd for all $n(\in\mathbb{N})$ | A1* | Must have correct opening contradiction statement |

### Proof by Induction:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume true for $n=k$, substitute $n=k+1$ into $(n+1)^3-n^3$, multiply out to three-term quadratic e.g. $3k^2+9k+7$ | M1 | Condone arithmetical slips |
| $(k+1)^3 - k^3 + 6(k+1) = f(k)+6(k+1)$, which is odd $+$ even $=$ odd | A1 | |
| $n=1 \Rightarrow (1+1)^3-1^3=7$ (true) | dM1 | Condone arithmetical slips |
| States: true for $n=1$; if true for $n=k$ then true for $n=k+1$; therefore true for all $n(\in\mathbb{N})$ | A1* | All three induction conclusion elements required |