Section B [80 marks] Answer any five questions in this section. Each question carries 16 marks.

$$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$$ $$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$ $$\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} \right]^2$$

So, the sum becomes:

$$\left[ \frac{n(n+1)}{2} \right]^2 + 2 \left[ \frac{n(n+1)(2n+1)}{6} \right] + \frac{n(n+1)}{2}$$ $$= \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$$

The common factor is $\frac{n(n+1)}{12}$.

$$= \frac{n(n+1)}{12} \left[ \frac{3n(n+1)}{1} + \frac{4(2n+1)}{1} + \frac{6}{1} \right]$$ $$= \frac{n(n+1)}{12} [3n^2 + 3n + 8n + 4 + 6]$$ $$= \frac{n(n+1)}{12} [3n^2 + 11n + 10]$$

Now, factorise the quadratic $3n^2 + 11n + 10$. We look for two numbers that multiply to $3 \times 10 = 30$ and add to 11. These are 5 and 6.
$3n^2 + 6n + 5n + 10 = 3n(n+2) + 5(n+2) = (3n+5)(n+2)$.
So, the fully factorised form is:

$$\boxed{\frac{n(n+1)(n+2)(3n+5)}{12}}$$

Part (a)(ii) Hence find $\sum_{r=10}^{30} r(r+1)^2$.

For $n=30$:

$$S_{30} = \frac{30(30+1)(30+2)(3(30)+5)}{12}$$ $$S_{30} = \frac{30 \times 31 \times 32 \times (90+5)}{12}$$ $$S_{30} = \frac{30 \times 31 \times 32 \times 95}{12}$$ $$S_{30} = 5 \times 31 \times 8 \times 95$$ $$S_{30} = 120960 - 120960 + 117800$$ $$S_{30} = 117800$$

Let's re-calculate $5 \times 31 \times 8 \times 95$:
$5 \times 8 = 40$
$40 \times 31 = 1240$
$1240 \times 95 = 117800$

For $n=9$:

$$S_9 = \frac{9(9+1)(9+2)(3(9)+5)}{12}$$ $$S_9 = \frac{9 \times 10 \times 11 \times (27+5)}{12}$$ $$S_9 = \frac{9 \times 10 \times 11 \times 32}{12}$$ $$S_9 = 3 \times 10 \times 11 \times 8$$ $$S_9 = 30 \times 88$$ $$S_9 = 2640$$

$$\sum_{r=10}^{30} r(r+1)^2 = S_{30} - S_9 = 117800 - 2640$$ $$= \boxed{115160}$$

Part (b) Obtain the expansion of $e^x \cos x$ up to and including the term in $x^4$.

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + O(x^5)$$ $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + O(x^5)$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)$$ $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$$

$$e^x \cos x = \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} \right) \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} \right)$$

Term by term multiplication:
Constant term: $1 \times 1 = 1$
$x$ term: $x \times 1 = x$
$x^2$ terms: $1 \times (-\frac{x^2}{2}) + \frac{x^2}{2} \times 1 = -\frac{x^2}{2} + \frac{x^2}{2} = 0$
$x^3$ terms: $x \times (-\frac{x^2}{2}) + \frac{x^3}{6} \times 1 = -\frac{x^3}{2} + \frac{x^3}{6} = -\frac{3x^3}{6} + \frac{x^3}{6} = -\frac{2x^3}{6} = -\frac{x^3}{3}$
$x^4$ terms: $1 \times \frac{x^4}{24} + \frac{x^2}{2} \times (-\frac{x^2}{2}) + \frac{x^4}{24} \times 1 = \frac{x^4}{24} - \frac{x^4}{4} + \frac{x^4}{24} = \frac{x^4}{24} - \frac{6x^4}{24} + \frac{x^4}{24} = \frac{(1-6+1)x^4}{24} = -\frac{4x^4}{24} = -\frac{x^4}{6}$

$$e^x \cos x = \boxed{1 + x - \frac{x^3}{3} - \frac{x^4}{6}}$$

Part (c)(i) Use Taylor's theorem for $f(x) = \ln x$ about $x=1$.

$$f(x) = \ln x \implies f(1) = \ln(1) = 0$$ $$f'(x) = \frac{1}{x} \implies f'(1) = 1$$ $$f''(x) = -\frac{1}{x^2} \implies f''(1) = -1$$ $$f'''(x) = \frac{2}{x^3} \implies f'''(1) = 2$$ $$f^{(4)}(x) = -\frac{6}{x^4} \implies f^{(4)}(1) = -6$$

The Taylor series expansion about $x=a$ is $f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f^{(4)}(a)}{4!}(x-a)^4 + \dots$
For $a=1$:

$$\ln x = 0 + \frac{1}{1!}(x-1) + \frac{-1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 + \frac{-6}{4!}(x-1)^4 + \dots$$ $$\ln x = (x-1) - \frac{1}{2}(x-1)^2 + \frac{2}{6}(x-1)^3 - \frac{6}{24}(x-1)^4 + \dots$$ $$\ln x = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \dots$$

The first four non-zero terms are:

$$\boxed{(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4}$$

Part (c)(ii) Hence obtain an approximation to $\ln(1.2)$.

If $x=1.2$, then $(x-1) = 0.2$.

$$\ln(1.2) \approx (0.2) - \frac{1}{2}(0.2)^2 + \frac{1}{3}(0.2)^3 - \frac{1}{4}(0.2)^4$$

$$(0.2) = 0.2$$ $$-\frac{1}{2}(0.2)^2 = -\frac{1}{2}(0.04) = -0.02$$ $$\frac{1}{3}(0.2)^3 = \frac{1}{3}(0.008) = \frac{0.008}{3} \approx 0.0026666...$$ $$-\frac{1}{4}(0.2)^4 = -\frac{1}{4}(0.0016) = -0.0004$$

$$\ln(1.2) \approx 0.2 - 0.02 + 0.0026666... - 0.0004$$ $$\ln(1.2) \approx 0.18 + 0.0026666... - 0.0004$$ $$\ln(1.2) \approx 0.1826666... - 0.0004$$ $$\ln(1.2) \approx 0.1822666...$$

Rounding to four decimal places:

$$\ln(1.2) \approx \boxed{0.1823}$$