## Question 7(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: $lhs = \frac{1}{1(2)} = \frac{1}{2}$, $rhs = \frac{1}{1+1} = \frac{1}{2}$, so true for $n=1$ | B1 | Shows result holds for $n=1$; minimum LHS $= \frac{1}{1(2)}$, RHS $= \frac{1}{1+1}$ |
| Assume true for $n=k$ so $\sum_{r=1}^{k} \frac{1}{r(r+1)} = \frac{k}{k+1}$ | M1 | Makes statement assuming result true for some value of $n$, say $k$ |
| $\sum_{r=1}^{k+1} \frac{1}{r(r+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)}$ | M1 | Attempts to add next term and makes progress by attempting common denominator |
| $= \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$ cso | A1 | Correct expression in terms of $k+1$, no errors seen |
| If true for $n=k$ then shown true for $n=k+1$; true for $n=1$, therefore true for all positive integers $n$ | A1 | Correct conclusion conveying all four bold ideas; not dependent on B1 |

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## Question 7(ii):

### Way 1: $f(k+1)$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: $3^{2n+4} - 2^{2n} = 3^6 - 2^2 = 729 - 4 = 725$, true for $n=1$ as 725 divisible by 5 | B1 | Shows result holds for $n=1$ |
| Assume true for $n=k$ so $3^{2k+4} - 2^{2k}$ divisible by 5 | M1 | Makes statement assuming result true for some $k$ |
| $f(k+1) = 3^{2(k+1)+4} - 2^{2(k+1)} = 3^{2k+6} - 2^{2k+2}$; look for $A \times 3^{2k+4} - A \times 2^{2k} + B \times 2^{2k}$ or $A \times 3^{2k+4} - A \times 2^{2k} + B \times 3^{2k+4}$ | M1 | Attempts $f(k+1)$ and attempts to express in terms of $f(k)$ |
| $= 9 \times 3^{2k+4} - 9 \times 2^{2k} + 5 \times 2^{2k}$ or $4 \times 3^{2k+4} - 4 \times 2^{2k} + 5 \times 3^{2k+4}$ $= 9f(k) + 5 \times 2^{2k}$ or $= 4f(k) + 5 \times 3^{2k+4}$ | A1 | Correct expression in terms of $f(k)$ |
| If true for $n=k$ then shown true for $n=k+1$; true for $n=1$, therefore true for all positive integers $n$ | A1 | Correct conclusion; not dependent on B1 |

### Way 2: $f(k+1) - f(k)$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: $3^{2n+4} - 2^{2n} = 725$, divisible by 5 | B1 | Shows result holds for $n=1$ |
| Assume true for $n=k$ so $3^{2k+4} - 2^{2k}$ divisible by 5 | M1 | |
| $f(k+1) - f(k) = 3^{2k+6} - 2^{2k+2} - 3^{2k+4} + 2^{2k}$; look for $A \times 3^{2k+4} - A \times 2^{2k} + B \times 2^{2k}$ or $A \times 3^{2k+4} - A \times 2^{2k} + B \times 3^{2k+4}$ $= 8 \times 3^{2k+4} - 8 \times 2^{2k} + 5 \times 2^{2k}$ or $3 \times 3^{2k+4} - 3 \times 2^{2k} + 5 \times 3^{2k+4}$ | M1 | Attempts $f(k+1)-f(k)$ and expresses in terms of $f(k)$ |
| $f(k+1) = 9f(k) + 5 \times 2^{2k}$ or $f(k+1) = 4f(k) + 5 \times 3^{2k+4}$ | A1 | Correct expression for $f(k+1)$ in terms of $f(k)$ |
| Correct conclusion as above | A1 | |

### Way 3: $f(k) = 5M$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: result true as 725 divisible by 5 | B1 | |
| Assume true for $n=k$ so $3^{2k+4} - 2^{2k} = 5M$ divisible by 5 | M1 | |
| $f(k+1) = 3^2 \times 3^{2k+4} - 2^2 \times 2^{2k} = 3^2(5M + 2^{2k}) - 2^2 \times 2^{2k}$ OR $= 3^2 \times 3^{2k+4} - 2^2 \times (3^{2k+4} - 5M)$ | M1 | Attempts $f(k+1)$ and writes in terms of $5M$ |
| $f(k+1) = 45M + 5 \cdot 2^{2k}$ OR $f(k+1) = 5 \cdot 3^{2k+4} + 20M$ | A1 | Correct expression in terms of $M$ and $2^{2k}$ or $M$ and $3^{2k+4}$ |
| Correct conclusion as above | A1 | |

### Way 4: $f(k+1) - mf(k)$

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1$: result true as 725 divisible by 5 | B1 | |
| Assume true for $n=k$ so $3^{2k+4} - 2^{2k}$ divisible by 5 | M1 | |
| $f(k+1) - mf(k) = 3^{2k+6} - 2^{2k+2} - m(3^{2k+4} - 2^{2k})$ $= (9-m) \cdot 3^{2k+4} - 4 \cdot 2^{2k} + m \cdot 2^{2k}$ $= (9-m)(3^{2k+4} - 2^{2k}) + 5 \cdot 2^{2k}$ | M1 | Attempts $f(k+1) - mf(k)$ and expresses in terms of $f(k)$ |
| $f(k+1) = (9-m) \cdot f(k) + 5 \cdot 2^{2k} + mf(k)$; note $m=4$ leads to $5 \cdot 3^{2k+4}$ | A1 | Correct expression for $f(k+1)$ in terms of $f(k)$ |
| Correct conclusion as above | A1 | |

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