# Question 3(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{(r+1)(r+3)} = \frac{A}{r+1} + \frac{B}{r+3}$, giving $2 = A(r+3)+B(r+1)$ | M1 | Any valid method for partial fractions |
| $\frac{2}{(r+1)(r+3)} = \frac{1}{r+1} - \frac{1}{r+3}$ | A1 | Award M1A1 if correct with no working (cover-up method acceptable) |

**(2 marks)**

# Question 3(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum\frac{2}{(r+1)(r+3)} = \left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\cdots+\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$ | M1A1ft | M1: list ≥3 terms at start and 2 at end showing cancellation, starting $r=1$, ending $r=n$ |
| $= \frac{1}{2}+\frac{1}{3}-\frac{1}{n+2}-\frac{1}{n+3}$ | | A1ft: correct remaining terms following through their PFs |
| $= \frac{5(n+2)(n+3)-6(n+3)-6(n+2)}{6(n+2)(n+3)}$ | M1 | For combining into single fraction (numerator at least 2 terms correct) |
| $= \frac{n(5n+13)}{6(n+2)(n+3)}$ $*$ | A1cso | Check all steps correct, particularly 3rd line from end |

**(4 marks)**

# Question 3(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{10}^{100} = \sum_{1}^{100} - \sum_{1}^{9}$ | M1 | |
| $= \frac{100(513)}{6\times102\times103} - \frac{9\times58}{6\times11\times12} = \frac{1425}{1751} - \frac{29}{44}$ | | |
| $= 0.1547\ldots = 0.155$ | A1 | |

**(2 marks)**

**Total: 8 Marks**

## Question 3c:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempting $\sum_{1}^{100} - \sum_{1}^{9}$ using the result from (b) with numbers substituted | M1 | Use of $\sum_{1}^{100} - \sum_{1}^{10}$ scores M0 |
| Sum $= 0.155$ | A1cso | |

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## Question 4a:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}\frac{d^2y}{dx^2} + y\frac{d^3y}{dx^3} + 2\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + 5\frac{dy}{dx} = 0$ | M1M1 | M1 for product rule on $y\frac{d^2y}{dx^2}$; M1 for differentiating $5y$ and using product/chain rule on $\left(\frac{dy}{dx}\right)^2$ |
| $\frac{d^3y}{dx^3} = \frac{-5\frac{dy}{dx} - 3\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2}}{y}$ | A2,1,0 | A1A1 if fully correct; A1A0 if one error; A0A0 if more than one error. Two sign errors and no other error: A1A0. Do NOT deduct if two $\frac{d^2y}{dx^2}$ terms shown separately |

**Alternative 1:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = \frac{-5y - \left(\frac{dy}{dx}\right)^2}{y} = -5 - \frac{1}{y}\left(\frac{dy}{dx}\right)^2$ | M1M1 | M1M1 for differentiating using product and chain rule (or quotient rule) |
| $\frac{d^3y}{dx^3} = \frac{1}{y^2}\left(\frac{dy}{dx}\right)^3 - \frac{2}{y}\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx^2}\right)$ | A2,1,0 | A1A1 fully correct; A1A0 one error; A0A0 more than one error |

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## Question 4b:

| Working/Answer | Mark | Guidance |
|---|---|---|
| When $x=0$: $\frac{dy}{dx}=2$, $y=2$; $\frac{d^2y}{dx^2} = \frac{1}{2}(-10-4) = -7$ | M1A1 | M1 for substituting $\frac{dy}{dx}=2$ and $y=2$ into equation to get numerical $\frac{d^2y}{dx^2}$ |
| $\frac{d^3y}{dx^3} = \frac{-10 - 3\times 2\times -7}{2} = 16$ | A1 | |
| $y = 2 + 2x - \frac{7}{2!}x^2 + \frac{16}{3!}x^3 + \ldots$ | M1 | For using $y = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0)+\ldots$ |
| $y = 2 + 2x - \frac{7}{2}x^2 + \frac{8}{3}x^3$ | A1 | Must have $y=\ldots$ in ascending powers of $x$ |

**Alternative to Q4b:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Set $y = 2 + 2x + ax^2 + bx^3$ | M1 | |
| $(2+2x+ax^2+bx^3)(2a+6bx)+(2+2ax+3bx^2\ldots)^2 + 5(2+2x+ax^2+bx^3)=0$ | M1 | |
| Coeffs $x^0$: $4a+4+10=0 \Rightarrow a = -\frac{7}{2}$ | A1 | |
| Coeffs $x$: $4a+12b+8a+10=0 \Rightarrow b = \frac{8}{3}$ | A1 | |
| $y = 2 + 2x - \frac{7}{2}x^2 + \frac{8}{3}x^3$ | A1 | |

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## Question 5a:

| Working/Answer | Mark | Guidance |
|---|---|---|
| I.F. $= e^{\int 2\tan x\,dx} = e^{2\ln\sec x} = \sec^2 x$ | M1A1 | |
| $y\sec^2 x = \int \sec^2 x \sin 2x\,dx$ | M1 | |
| $y\sec^2 x = \int \frac{2\sin x\cos x}{\cos^2 x}dx = 2\int\tan x\,dx$ | | |
| $y\sec^2 x = 2\ln\sec x\,(+c)$ | M1depA1 | |
| $y = \frac{2\ln\sec x + c}{\sec^2 x}$ | A1ft | **(6)** |

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## Question 5b:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Substituting $y=2$, $x=\frac{\pi}{3}$: $2 = \frac{2\ln(2)+c}{4}$, so $c = 8 - 2\ln 2$ | M1A1 | |
| $x=\frac{\pi}{6}$: $y = \frac{2\ln\sec\!\left(\frac{\pi}{6}\right)+8-2\ln 2}{\sec^2\!\left(\frac{\pi}{6}\right)} = \frac{2\ln\frac{2}{\sqrt{3}}+8-2\ln 2}{\frac{4}{3}}$ | M1 | |
| $y = \frac{3}{4}\!\left(8 + 2\ln\frac{1}{\sqrt{3}}\right) = 6 + \frac{3}{2}\ln\frac{1}{\sqrt{3}} = 6 - \frac{3}{4}\ln 3$ | A1 | **(4)** |