# Question 6:

## Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$: $\sum_{r=1}^{1}(3r+1)(r+2) = 4\times3 = 12$ and $1(1+2)(1+3)=12$, true for $n=1$ | B1 | Verifies base case |
| Assume true for $n=k$: $\sum_{r=1}^{k}(3r+1)(r+2) = k(k+2)(k+3)$ | M1 | States assumption for $n=k$ |
| $\sum_{r=1}^{k+1}(3r+1)(r+2) = k(k+2)(k+3) + (3k+4)(k+3)$ | M1 | Adds $(k+1)$th term to assumed sum |
| $= (k+3)(k^2+5k+4)$ | A1 | Correct factorisation |
| $= (k+1)(k+3)(k+4)$ | A1 | Correct form matching $n=k+1$ |
| $= (k+1)(k+1+2)(k+1+3)$; conclusion: true for $n=k \Rightarrow$ true for $n=k+1$, true for $n=1$, therefore true for all positive integers $n$ | A1 | Correct conclusion with all elements |

## Part (ii) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$: $4^1+5^1+6^1=15$, divisible by 15, true for $n=1$ | B1 | Verifies base case |
| Assume true for $n=k$: $4^k+5^k+6^k$ divisible by 15 | M1 | States assumption |
| $f(k+2) = 4^{k+2}+5^{k+2}+6^{k+2}$ | M1 | Considers $f(k+2)$ |
| $= 16\times4^k+16\times5^k+16\times6^k+9\times5^k+20\times6^k$ | A1 | Correct expansion |
| $= 16f(k)+45\times5^{k-1}+120\times6^{k-1}$ | A1 | Written in terms of $f(k)$ |
| Since 15 divides $f(k)$, 45 and 120, 15 divides $f(k+1)$; true for $n=1$, true for $n=k \Rightarrow$ true for $n=k+2$, true for all positive odd integers $n$ | A1 | Correct conclusion |

## Part (ii) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$: $4^1+5^1+6^1=15$, true for $n=1$ | B1 | Verifies base case |
| Assume true for $n=k$: $4^k+5^k+6^k$ divisible by 15 | M1 | States assumption |
| $f(k+2)-f(k) = 4^{k+2}+5^{k+2}+6^{k+2}-4^k-5^k-6^k$ | M1 | Considers difference |
| $= 15\times4^k+24\times5^k+35\times6^k = 15f(k)+45\times5^{k-1}+120\times6^{k-1}$ | A1 | Correct simplification |
| $f(k+2) = 16f(k)+45\times5^{k-1}+120\times6^{k-1}$ | A1 | Correct expression for $f(k+2)$ |
| $f(k+2) = 16f(k)+15(3\times5^{k-1}+8\times6^{k-1})$; true for $n=k \Rightarrow$ true for $n=k+2$, true for $n=1$, true for all positive odd integers $n$ | A1 | Correct conclusion |

## Question 6(ii) Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $n=1$, $4^1 + 5^1 + 6^1 = 15$, so statement true for $n=1$ | B1 | Shows statement true for $n=1$ by evaluating both sides |
| Assume true for $n=2k+1$ so $f(2k+1) = 4^{2k+1} + 5^{2k+1} + 6^{2k+1}$ is divisible by 15 | M1 | Makes assumption statement that result is true for some odd value of $n$ |
| $f(2k+3) = 4^{2k+3} + 5^{2k+3} + 6^{2k+3}$ or $f(2k+3) - f(2k+1) = 4^{2k+3}+5^{2k+3}+6^{2k+3}-4^{2k+1}-5^{2k+1}-6^{2k+1}$ | M1 | Attempts $f(2k+3)$ or $f(2k+3)-f(2k+1)$ |
| $f(2k+3) = 16\times 4^{2k+1}+25\times 5^{2k+1}+36\times 6^{2k+1}$ $= 16(4^{2k+1}+5^{2k+1}+6^{2k+1})+9\times 5^{2k+1}+20\times 6^{2k+1}$ OR $f(2k+3)-f(2k+1) = 15\times 4^{2k+1}+120\times 5^{2k}+210\times 6^{2k}$ | A1 | Reaches $16(4^{2k+1}+5^{2k+1}+6^{2k+1})+9\times 5^{2k+1}+20\times 6^{2k+1}$ or suitable equivalent |
| $f(2k+3) = 16(4^{2k+1}+5^{2k+1}+6^{2k+1})+45\times 5^{2k}+120\times 6^{2k}$ OR $f(2k+3) = f(2k+1)+15\times 4^{2k+1}+120\times 5^{2k}+210\times 6^{2k}$ | A1 | Correct expression of form $Af(2k+1)+15(\ldots)$ |
| $f(2k+3) = 16(4^{2k+1}+5^{2k+1}+6^{2k+1})+15(3\times 5^{2k}+8\times 6^{2k})$ OR $f(2k+3) = f(2k+1)+15(4^{2k+1}+8\times 5^{2k}+14\times 6^{2k})$ and: If true for $n=2k+1$ then true for $n=2k+3$, true for $n=1$ so true for all positive odd integers $n$ | A1 | Correct conclusion with all four underlined ideas conveyed; do not allow "true for all $n$" where $n$ represents natural numbers |

---