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1995 Paper 1 Q3
D: 1500.0 B: 1500.0
sequences telescoping sum summation cube numbers method of differences alternating series polynomial identity

  1. If \(\mathrm{f}(r)\) is a function defined for \(r=0,1,2,3,\ldots,\) show that \[ \sum_{r=1}^{n}\left\{ \mathrm{f}(r)-\mathrm{f}(r-1)\right\} =\mathrm{f}(n)-\mathrm{f}(0). \]
  2. If \(\mathrm{f}(r)=r^{2}(r+1)^{2},\) evaluate \(\mathrm{f}(r)-\mathrm{f}(r-1)\) and hence determine \({\displaystyle \sum_{r=1}^{n}r^{3}.}\)
  3. Find the sum of the series \(1^{3}-2^{3}+3^{3}-4^{3}+\cdots+(2n+1)^{3}.\)

Solution:

  1. \(\,\) \begin{align*} && \sum_{r=1}^n \left (f(r) - f(r-1) \right) &= \cancel{f(1)} - f(0) + \cdots\\ &&&\quad\, \cancel{\f(2)}-\cancel{f(1)} + \cdots \\ &&&\quad\, \cancel{f(3)}-\cancel{f(2)} + \cdots \\ &&&\quad\, +\cdots + \cdots \\ &&&\quad\, \cancel{f(n-1)}-\cancel{f(n-2)} + \cdots \\ &&&\quad\, f(n)-\cancel{f(n-1)} \\ &&&=f(n) - f(0) \end{align*}
  2. If \(f(r) = r^2(r+1)^2\) then \begin{align*} && f(r) - f(r-1) &= r^2(r+1)^2 - (r-1)r^2 \\ &&&= r^2((r+1)^2-(r-1)^2) \\ &&&=4r^3 \end{align*} Therefore \begin{align*} && \sum_{r=1}^n 4r^3 &= n^2(n+1)^2-0 \\ \Rightarrow && \sum_{r=1}^n r^3 &= \frac{n^2(n+1)^2}{4} \\ \end{align*}
  3. So \begin{align*} && \sum_{r=1}^{2n+1} (-1)^{r-1}r^3 &= \sum_{r=1}^{2n+1} r^3 - 2 \sum_{r=1}^n (2r)^3 \\ &&&= \frac14(2n+1)^2(2n+2)^2 - 8 \frac{n^2(n+1)^2}{4} \\ &&&= (2n+1)^2(n+1)^2 - 2n^2(n+1)^2 \\ &&&= (n+1)^2(4n^2+4n+1 - 2n^2) \\ &&&= (n+1)^2(2n^2+4n+1) \end{align*}

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1995 Paper 2 Q1
D: 1600.0 B: 1484.0
geometric series differentiation alternating series polynomial identity summation substitution x=-1 second derivative

  1. By considering \((1+x+x^{2}+\cdots+x^{n})(1-x)\) show that, if \(x\neq1\), \[ 1+x+x^{2}+\cdots+x^{n}=\frac{1-x^{n+1}}{1-x}. \]
  2. By differentiating both sides and setting \(x=-1\) show that \[ 1-2+3-4+\cdots+(-1)^{n-1}n \] takes the value \(-n/2\) is \(n\) is even and the value \((n+1)/2\) if \(n\) is odd.
  3. Show that \[ 1^{2}-2^{2}+3^{2}-4^{2}+\cdots+(-1)^{n-1}n^{2}=(-1)^{n-1}(An^{2}+Bn) \] where the constants \(A\) and \(B\) are to be determined.

Solution:

  1. \begin{align*} && (1+x+x^{2}+\cdots+x^{n})(1-x) &= 1-x+x-x^2+\cdots -x^n+x^n-x^{n+1} \\ &&&= 1-x^{n+1} \\ \Rightarrow && 1+x+x^2+\cdots+x^n &= \frac{1-x^{n+1}}{1-x} \tag{dividing by \(1-x\)} \end{align*}
  2. \begin{align*} \frac{\d}{\d x}: && 0+1+2x+\cdots+nx^{n-1} &= \frac{(n+1)(1-x)x^n+(1-x^{n+1})}{(1-x)^2} \\ \Rightarrow && 1-2x+\cdots+(-1)^n n &= \frac{-(n+1)2(-1)^n+(1-(-1)^{n+1})}{4} \\ &&&= \begin{cases} \frac{-(n+1)\cdot2\cdot1+(1-(-1)}{4} & \text{if }n\text{ is even} \\ \frac{-(n+1)\cdot 2 \cdot(-1)+(1-1)}{4} & \text{if }n\text{ is odd}\end{cases} \\ &&&= \begin{cases} \frac{-n}{2} & \text{if }n\text{ is even}\\ \frac{n+1}{2} & \text{if }n\text{ is odd}\end{cases} \\ \end{align*}
  3. \begin{align*} x: && x+2x^2+\cdots+nx^{n} &= \frac{(n+1)(1-x)x^{n+1}+x(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{x+(n+1)x^{n+1}-nx^{n+2}}{(1-x)^2}\\ \frac{\d}{\d x}: && 1^2+2^2x + \cdots + n^2x^{n-1} &= \frac{(1-x)^2(1+(n+1)^2x^{n}-n(n+2)x^{n+1}) +2(1-x)(x+(n+1)x^{n+1}-nx^{n+2})}{(1-x)^4} \\ &&&= \frac{1 + x - (1 + n)^2 x^n + (2 n^2+2n-1) x^{n+1} - n^2 x^{n+2}}{(1-x)^3} \\ \Rightarrow && 1^2-2^2 + \cdots + (-1)^{n-1}n^2 &= \frac{(-1)^n \l - (1 + n)^2- (2 n^2+2n-1) - n^2 \r}{8} \\ &&&= \frac{(-1)^n(-4n^2-4n}{8} \\ &&&= \frac{(-1)^{n-1}(n^2+n)}{2} \end{align*}

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1991 Paper 2 Q5
D: 1600.0 B: 1516.0
reduction formulae integration curve sketching tan powers series alternating series ln 2 Leibniz formula for pi

Give a rough sketch of the function \(\tan^{k}\theta\) for \(0\leqslant\theta\leqslant\frac{1}{4}\pi\) in the two cases \(k=1\) and \(k\gg1\) (i.e. \(k\) is much greater than 1). Show that for any positive integer \(n\) \[ \int_{0}^{\frac{1}{4}\pi}\tan^{2n+1}\theta\,\mathrm{d}\theta=(-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right), \] and deduce that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m}=\tfrac{1}{2}\ln2. \] Show similarly that \[ \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m-1}=\frac{\pi}{4}. \]

Solution:

TikZ diagram
Let \(\displaystyle I_n = \int_0^{\pi/4} \tan^{n} \theta \, \d \theta\), then \begin{align*} I_0 &= \int_0^{\pi/4} \tan \theta \d \theta \\ &= \left [ -\ln \cos \theta \right]_0^{\pi/4} \\ &= -\ln \frac{1}{\sqrt{2}} - 0 \\ &= \frac12 \ln 2 \\ \\ \\ I_{2n+1} &= \int_0^{\pi/4} \tan^{2n+1} \theta \, \d \theta \\&= \int_0^{\pi/4} \tan^{2n-1} \theta \tan ^2 \theta \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta (\sec^2 \theta - 1) \, \d \theta \\ &= \int_0^{\pi/4} \tan^{2n-1} \theta \sec^2 \theta - \tan^{2n-1} \theta \, \d \theta \\ &= \left[ \frac{1}{2n} \tan^{2n} \theta \right]_0^{\pi/4} - I_{2n-1} \\ &= \frac1{2n} - I_{2n-1} \end{align*} Therefore we can see that \(\displaystyle I_{2n+1} = (-1)^{n}\left(\tfrac{1}{2}\ln2+\sum_{m=1}^{n}\frac{(-1)^{m}}{2m}\right)\). As we can see as \(n \to \infty\), \(I_n \to 0\) Therefore \begin{align*} && 0 &= \tfrac{1}{2}\ln2+\lim_{n \to \infty} \sum_{m=1}^{n}\frac{(-1)^{m}}{2m} \\ \Rightarrow && \tfrac{1}{2}\ln2 &= \sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{2m} \end{align*} \begin{align*} && I_{-1} &= \int_0^{\pi/4} 1 \d \theta \\ &&&= \frac{\pi}{4} \end{align*} Therefore \(\displaystyle I_{2n} = (-1)^n \left ( \frac{\pi}{4} + \sum_{m=1}^n \frac{(-1)^m}{2m-1} \right)\) and since \(I_{2m} \to 0\) the same result follows.

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