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1991 Paper 3 Q8
D: 1700.0 B: 1500.1
reduction formula integration by parts trigonometric binomial coefficients cos nθ summation proof

  1. The integral \(I_{k}\) is defined by \[ I_{k}=\int_{0}^{\theta}\cos^{k}x\,\cos kx\,\mathrm{d}x. \] Prove that \(2kI_{k}=kI_{k-1}+\cos^{k}\theta\sin k\theta.\)
  2. Prove that \[ 1+m\cos2\theta+\binom{m}{2}\cos4\theta+\cdots+\binom{m}{r}\cos2r\theta+\cdots+\cos2m\theta=2^{m}\cos^{m}\theta\cos m\theta. \]
  3. Using the results of (i) and (ii), show that \[ m\frac{\sin2\theta}{2}+\binom{m}{2}\frac{\sin4\theta}{4}+\cdots+\binom{m}{r}\frac{\sin2r\theta}{2r}+\cdots+\frac{\sin2m\theta}{2m} \] is equal to \[ \cos\theta\sin\theta+\cos^{2}\theta\sin2\theta+\cdots+\frac{1}{r}2^{r-1}\cos^{r}\theta\sin r\theta+\cdots+\frac{1}{m}2^{m-1}\cos^{m}\theta\sin m\theta. \]

Solution:

  1. \begin{align*} kI_k &= \int_0^\theta \cos^k x k\cos kx \d x \\ &= \left [\cos^k x \sin kx \right]_0^\theta - \int_0^\theta k \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta - k\int_0^\theta \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta + k\int_0^\theta \cos^{k-1} x (\cos (k-1) x - \cos k x \cos x) \d x \\ &= \cos^k\theta \sin k \theta + k I_{k-1} - kI_k \\ \end{align*} So \(2kI_k = kI_{k-1} + \cos^k \sin k \theta\)
  2. \begin{align*} && \sum_{r=0}^m \binom{m}{r} \cos 2r \theta &= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2r \theta)\right) \\ &&&= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2 \theta)^r\right) \\ &&&= \textrm{Re} \left ( \left (1+e^{2i \theta} \right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}(e^{i \theta} +e^{-i\theta})\right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}2\cos \theta\right)^m\right) \\ &&&= \textrm{Re} \left (2^m \cos^m \theta e^{im\theta}\right) \\ &&&= 2^m \cos^m \theta \cos m\theta \\ \end{align*}
  3. \begin{align*} \sum_{r=1}^m \binom{m}{r}\frac{\sin 2r \theta}{2r} &= \int_0^\theta \sum_{r=1}^m \binom{m}{r}\cos 2r x\d x \\ &= \int_0^\theta \left (2^m \cos^m x\cos mx- 1 \right ) \d x\\ &= 2^m I_m - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + 2^{m-1} I_{m-1} - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + \frac{2^{m-2}}{m-1}\cos^m \theta \sin m \theta+2^{m-2} I_{m-2} - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + I_0 - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + \int_0^\theta 1 \cdot 1 \d \theta - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta \\ &= \sum_{r=0}^{m} \frac{2^{r-1}}{r} \cos^{r} \theta \sin r \theta \end{align*}

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1990 Paper 1 Q2
D: 1500.0 B: 1516.0
complex numbers De Moivre roots of unity binomial theorem binomial coefficients trig identity modulus and argument cube roots of unity

Let \(\omega=\mathrm{e}^{2\pi\mathrm{i}/3}.\) Show that \(1+\omega+\omega^{2}=0\) and calculate the modulus and argument of \(1+\omega^{2}.\) Let \(n\) be a positive integer. By evaluating \((1+\omega^{r})^{n}\) in two ways, taking \(r=1,2\) and \(3\), or otherwise, prove that \[ \binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots+\binom{n}{k}=\frac{1}{3}\left(2^{n}+2\cos\left(\frac{n\pi}{3}\right)\right), \] where \(k\) is the largest multiple of \(3\) less than or equal to \(n\). Without using a calculator, evaluate \[ \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} \] and \[ \binom{24}{2}+\binom{24}{5}+\cdots+\binom{24}{23}\,. \] {[}\(2^{25}=33554432.\){]}

Solution: Since \(\omega^3 = 1\) and \(\omega \neq 1\) we must have that \((\omega-1)(1 + \omega + \omega^2) = 0\) but by dividing by \(\omega - 1\) we obtain the desired result. \(1+\omega^2 = -\omega\) so \(|1 + \omega^2| = |-\omega| = 1\) and \(\arg ( 1 + \omega^2) = \arg(-\omega) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}\) \begin{align*} && (1 + 1)^n &= \sum_{k=0}^n \binom{n}{k}\\ && (1+ \omega)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{k} \\ && (1+ \omega^2)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{2k} \\ \Rightarrow && 2^n+(-\omega^2)^n + (-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n (1+1+1)\binom{n}{k} + \sum_{k=0, k \equiv 1 \pmod{3}}^n (1 + \omega + \omega^2) \binom{n}{k} + \sum_{k=0, k \equiv 2 \pmod{3}}^n (1 + \omega^2 + \omega) \binom{n}{k} \\ \Rightarrow && 2^n +((-\omega)^n)^{-1}+(-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n \binom{n}{k} \end{align*} \(2^n +((-\omega)^n)^{-1}+(-\omega)^n = 2^n + 2 \textrm{Re}(-\omega^n) = 2^n + 2 \cos \frac{n\pi}{3}\) Therefore our answer follows. \begin{align*} \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} &= \frac13 \l 2^{25} + 2\cos (\frac{25 \pi}{3}) \r \\ &= \frac13 \l 2^{25} + 2 \cos \frac{\pi}{3} \r \\ &= \frac13 \l 2^{25} + 1 \r \\ &= \frac13 \l (4096 \cdot 4096 \cdot 2) + 1 \r \\ &= 11\,184\,811 \end{align*} Notice that \(S_2 = \binom{24}{2} + \cdots +\binom{24}{23} = \binom{24}{1} + \cdots + \binom{24}{22} = S_1\) and \(S_0 = \binom{24}0 + \cdots + \binom{24}{21} = \frac13 \l 2^{24} + 2 \r\) Therefore since \(S_0 + 2 \cdot S_2 = 2^{24}\) we must have \begin{align*} S_2 &= \frac12 \l 2^{24} - \frac13 \l 2^{24} + 2 \r \r \\ &= \frac13 \l 2^{24} - 1 \r \\ &= \frac13 \l 16777216- 1 \r \\ &= \frac13 \cdot 16777215 \\ &= 5\,592\,405 \end{align*}

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1990 Paper 3 Q5
D: 1700.0 B: 1500.0
differentiation Leibniz rule proof by induction binomial coefficients Pascal's rule nth derivative inverse trigonometric functions recurrence relation

Prove that, for any integers \(n\) and \(r\), with \(1\leqslant r\leqslant n,\) \[ \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}. \] Hence or otherwise, prove that \[ (uv)^{(n)}=u^{(n)}v+\binom{n}{1}u^{(n-1)}v^{(1)}+\binom{n}{2}u^{(n-2)}v^{(2)}+\cdots+uv^{(n)}, \] where \(u\) and \(v\) are functions of \(x\) and \(z^{(r)}\) means \(\dfrac{\mathrm{d}^{r}z}{\mathrm{d}x^{r}}\). Prove that, if \(y=\sin^{-1}x,\) then \((1-x^{2})y^{(n+2)}-(2n+1)xy^{(n+1)}-n^{2}y^{(n)}=0.\)

Solution: \begin{align*} \binom{n}{r} + \binom{n}{r-1} &= \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \\ &= \frac{n!}{(r-1)!(n-r)!} \left ( \frac{1}{r} + \frac{1}{n-r+1} \right) \\ &= \frac{n!}{(r-1)!(n-r)!} \frac{(n-r+1)+r}{r(n-r+1)} \\ &= \frac{n! (n+1)}{r! (n-r+1)!} \\ &= \frac{(n+1)!}{r!(n+1-r)!} \\ &= \binom{n+1}{r} \end{align*} Claim: \(\displaystyle (uv)^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}\) Proof: (By induction on \(n\)). Base case: \(n = 0\) is clear. Inductive step: Suppose it is true for \(n = k\), then consider \begin{align*} (uv)^{(k+1)} &= \left ( (uv)^{(k)} \right)' \\ &= \left ( \sum_{r=0}^k \binom{k}{r} u^{(k-r)} v^{(r)} \right)' \tag{by assumption} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r)} v^{(r)}\right)' \tag{linearity} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r+1)} v^{(r)} + u^{(k-r)}v^{(r+1)}\right) \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=0}^{k} \binom{k}{r} u^{(k-r)}v^{(r+1)} \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=1}^{k+1} \binom{k}{r-1} u^{(k-r+1)}v^{(r)} \\ &= u^{(k+1)}v + \sum_{r=1}^k \left (\binom{k}{r} + \binom{k}{r-1} \right)u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= u^{(k+1)}v + \sum_{r=1}^k \binom{k+1}{r} u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= \sum_{r=0}^{k+1} \binom{k+1}{r} u^{(k-r+1)}v^{(r)}\\ \end{align*} Therefore if our statement is true for \(n = k\) it is true for \(n = k+1\). Since it is true for \(n = 0\) by the principle of mathematical induction it is true for all integer \(n \geq 0\) Suppose \( y = \sin^{-1} x\), then \(y' = \frac{1}{\sqrt{1-x^2}}\), \(y'' = \frac{x}{(1-x^2)^{3/2}}\). Not that this means that \((1-x^2)y'' - xy' = 0\) (which is our formula when \(n = 0\)). Now apply Leibniz's formula to this. \begin{align*} 0 &= \left ( (1-x^2)y'' - xy' \right)^{(n)} \\ &= \left ( (1-x^2)y'' \right)^{(n)} -\left ( xy' \right)^{(n)} \\ &= \left ( (1-x^2)y^{(n+2)} - n\cdot 2x \cdot y^{(n+1)}-\binom{n}{2} \cdot 2 \cdot y^{(n)} \right )- \left (xy^{(n+1)}+ny^{(n)} \right) \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - \left ( n(n-1)+n \right)y^{(n)} \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - n^2y^{(n)} \\ \end{align*} as required

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1989 Paper 1 Q5
D: 1500.0 B: 1516.0
binomial theorem binomial coefficients substitution summation even and odd terms differentiation positive integer n

Write down the binomial expansion of \((1+x)^{n}\), where \(n\) is a positive integer.

  1. By substituting particular values of \(x\) in the above expression, or otherwise, show that, if \(n\) is an even positive integer, \[ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{n}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots+\binom{n}{n-1}=2^{n-1}. \]
  2. Show that, if \(n\) is any positive integer, then \[ \binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots+n\binom{n}{n}=n2^{n-1}. \]
Hence evaluate \[ \sum_{r=0}^{n}\left(r+(-1)^{r}\right)\binom{n}{r}\,. \]

Solution: \[ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \]

  1. \begin{align*} (1+1)^n &= \sum_{k=0}^n \binom{n}{k} \\ (1-1)^n &= \sum_{k=0}^n (-1)^n\binom{n}{k} \\ &= \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} -\sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} \end{align*} Therefore \(\displaystyle \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} = \sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} = \frac{2^n}{2} = 2^{n-1}\)
  2. \begin{align*} && (1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{k=0}^n k\binom{n}{k} x^{k-1} \\ x = 1: && n2^{n-1} &= \sum_{k=1}^n k\binom{n}{k} \end{align*} as required
\begin{align*} \sum_{r=0}^n (r + (-1)^r) \binom{n}{r} &= n2^{n-1}+0 = n2^{n-1} \end{align*}

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1988 Paper 1 Q4
D: 1500.0 B: 1516.0
combinatorics counting proof line segments intersections average binomial coefficients

Each of \(m\) distinct points on the positive \(y\)-axis is joined by a line segment to each of \(n\) distinct points on the positive \(x\)-axis. Except at the endpoints, no three of these segments meet in a single point. Derive formulae for \begin{questionparts} \item the number of such line segments; \item the number of points of intersections of the segments, ignoring intersections at the endpoints of the segments. \end{questionpart} If \(m=n\geqslant3,\) and the two segments with the greatest number of points of intersection, and the two segments with the least number of points of intersection, are excluded, prove that the average number of points of intersection per segment on the remaining segments is \[ \frac{n^{3}-7n+2}{4(n+2)}\,. \]

Solution:

  1. There are \(m \cdot n\) line segments, since each of the \(m\) points on the \(y\)-axis correspond to \(n\) points of the \(x\)-axis.
  2. Every pair of points on the \(x\)-axis and pair of points on the \(y\) axis will generate \(4\) lines, but only \(1\) of those points will meet in the middle, therefore there will be \(\binom{m}{2} \binom{n}{2}\) intersections.
The segment joining the smallest \(x\) to the smallest \(y\) and the segment joining the largest \(x\) to the largest \(y\) will not meet any other segments. The segment joining the smallest \(x\) to the largest \(y\) and the segment joining the largest \(x\) to the smallest \(y\) will have the largest number of intersections. They each intersect all the other segment coming out of other points (except the ones headed for the same point) ie \((m-1)(n-1)\) each. They also intersect each other, so \(2(m-1)(n-1)-1\) points. Therefore we are interested in: \begin{align*} \frac{\binom{n}{2}\binom{n}{2} - 2(n-1)^2+1}{n^2-4} &= \frac{\frac{n^2(n-1)^2}{4} - 2n^2+4n-1}{n^2-4} \\ &= \frac{n^4-2n^3+n^2-8n^2+16n-4}{4(n^2-4)} \\ &= \frac{(n - 2) (n^3 - 7 n + 2)}{4(n^2-4)} \\ &= \frac{n^3-7n+3}{4(n+2)} \end{align*}

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