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2002 Paper 1 Q8
I borrow \(C\) pounds at interest rate \(100\alpha \,\%\) per year. The interest is added at the end of each year. Immediately after the interest is added, I make a repayment. The amount I repay at the end of the \(k\)th year is \(R_k\) pounds and the amount I owe at the beginning of \(k\)th year is \(C_k\) pounds (with \(C_1=C\)). Express \(C_{n+1}\) in terms of \(R_k\) (\(k= 1\), \(2\), \(\ldots\), \(n\)), \(\alpha\) and \(C\) and show that, if I pay off the loan in \(N\) years with repayments given by \(R_k= (1+\alpha)^kr\,\), where \(r\) is constant, then \(r=C/N\,\). If instead I pay off the loan in \(N\) years with \(N\) equal repayments of \(R\) pounds, show that \[ \frac R C = \frac{\alpha (1+\alpha)^{N} }{(1+\alpha)^N-1} \;, \] and that \(R/C\approx 27/103\) in the case \(\alpha =1/50\), \(N=4\,\).
2002 Paper 1 Q14
In order to get money from a cash dispenser I have to punch in an identification number. I have forgotten my identification number, but I do know that it is equally likely to be any one of the integers \(1\), \(2\), \ldots , \(n\). I plan to punch in integers in order until I get the right one. I can do this at the rate of \(r\) integers per minute. As soon as I punch in the first wrong number, the police will be alerted. The probability that they will arrive within a time \(t\) minutes is \(1-\e^{-\lambda t}\), where \(\lambda\) is a positive constant. If I follow my plan, show that the probability of the police arriving before I get my money is \[ \sum_{k=1}^n \frac{1-\e^{-\lambda(k-1)/r}}n\;. \] Simplify the sum. On past experience, I know that I will be so flustered that I will just punch in possible integers at random, without noticing which I have already tried. Show that the probability of the police arriving before I get my money is \[ 1-\frac1{n-(n-1)\e^{-\lambda/r}} \;. \]
2002 Paper 2 Q14
A densely populated circular island is divided into \(N\) concentric regions \(R_1\), \(R_2\), \(\ldots\,\), \(R_N\), such that the inner and outer radii of \(R_n\) are \(n-1\) km and \(n\) km, respectively. The average number of road accidents that occur in any one day in \(R_n\) is \(2-n/N\,\), independently of the number of accidents in any other region. Each day an observer selects a region at random, with a probability that is proportional to the area of the region, and records the number of road accidents, \(X\), that occur in it. Show that, in the long term, the average number of recorded accidents per day will be \[ 2-\frac16\left(1+\frac1N\right)\left(4-\frac1N\right)\;. \] [Note: \(\sum\limits_{n=1}^N n^2 = \frac16 N(N+1)(2N+1) \;\).] Show also that \[ \P(X=k) = \frac{\e^{-2}N^{-k-2}}{k!}\sum_{n=1}^N (2n-1)(2N-n)^k\e^{n/N} \;. \] Suppose now that \(N=3\) and that, on a particular day, two accidents were recorded. Show that the probability that \(R_2\) had been selected is \[ \frac{48}{48 + 45\e^{1/3} +25 \e^{-1/3}}\;. \]
Solution: The area of \(R_n\) is \(\pi(n^2 - (n-1)^2) = (2n-1)\pi\). The area of the whole region is \(\pi N^2\). \begin{align*} && \E[X] &= \E[\E[X | \text{choose region }n]] \\ &&&= \sum_{n=1}^N \left (2 - \frac{n}{N} \right) \cdot \frac{(2n-1)\pi}{N^2 \pi} \\ &&&= \sum_{n=1}^N \left (2\cdot \frac{(2n-1)\pi}{N^2 \pi} - \frac{n}{N}\cdot \frac{(2n-1)\pi}{N^2 \pi} \right) \\ &&&= 2 - \frac{1}{N^3} \sum_{n=1}^N (2n^2-n) \\ &&&= 2 - \frac{1}{N^3} \left (\frac{2N(N+1)(2N+1)}{6} - \frac{N(N+1)}{2} \right) \\ &&&= 2 - \frac{N+1}{6N^2} \left (2(2N+1)-3 \right) \\ &&&= 2 - \frac{N+1}{6N^2} (4N - 1) \\ &&&= 2 - \frac16 \left (1 + \frac1N \right) \left (4 - \frac1N \right) \end{align*} Modelling each region as \(Po(2 - n/N)\) we have \begin{align*} \mathbb{P}(X = k ) &= \sum_{n=1}^N \exp(-2 + n/N) \frac{(2-n/N)^k}{k!} \frac{2n-1}{N^2} \\ &= \frac{e^{-2}N^{-k-2}}{k!} \sum_{n=1}^N e^{n/N} (2N-n)^k(2n-1) \end{align*} as desired. Supposing \(N=3\) and two accidents then \begin{align*} \mathbb{P}(R_2 | X = 2) &= \frac{\frac{3}{9} e^{-4/3}\frac{(\frac43)^2}{2!}}{\mathbb{P}(X=2)} \\ &= \frac{\frac{3}{9} e^{-4/3} \frac{(\frac43)^2}{2!}}{\frac{1}{9} e^{-5/3} \frac{(\frac53)^2}{2!} + \frac{3}{9} e^{-4/3} \frac{(\frac43)^2}{2!} + \frac{5}{9} e^{-2/3} \frac{(\frac33)^2}{2!}} \\ &= \frac{3 \cdot 16}{25e^{-1/3} + 3 \cdot 16 + 5 \cdot 9e^{1/3}} \\ &= \frac{48}{25e^{-1/3} + 48 + 45e^{1/3}} \end{align*} as required.
2002 Paper 3 Q2
Prove that \(\displaystyle \arctan a + \arctan b = \arctan \l {a + b \over 1-ab} \r\,\) when \(0 < a < 1\) and \(0 < b < 1\,\). Prove by induction that, for \(n \ge 1\,\), \[ \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r \] and hence find \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 + r + 1} \r\,. \] Hence prove that \[ \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r = {\pi \over 2}\,. \]
Solution: \begin{align*} && \arctan a &\in (0, \tfrac{\pi}{4}) \\ && \arctan b &\in (0, \tfrac{\pi}{4}) \\ \Rightarrow && \arctan a+\arctan b &\in (0, \tfrac{\pi}{2}) \\ && \tan \left ( \arctan a+\arctan b \right) &= \frac{\tan \arctan a + \tan \arctan b}{1 - \tan \arctan a \tan \arctan b} \\ &&&= \frac{a+b}{1-ab} \in (0, \infty) \\ \Rightarrow && \arctan \left ( \frac{a+b}{1-ab} \right) &\in (0, \tfrac{\pi}{2}) \\ \Rightarrow && \arctan a + \arctan b &= \arctan \left ( \frac{a+b}{1-ab} \right) \end{align*} Claim: \(\displaystyle \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r = \arctan \l {n \over n+2} \r\) Proof: (By Induction): Base case (\(n=1\)): \begin{align*} && LHS &= \sum_{r=1}^1 \arctan \left ( \frac{1}{r^2+r+1} \right) \\ &&&= \arctan \left ( \frac{1}{3} \right) \\ && RHS &= \arctan \left ( \frac{1}{1+2} \right)\\ &&&= \arctan \left ( \frac{1}{3} \right) = LHS \end{align*} Inductive step, suppose true for \(n = k\), ie \begin{align*} && \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r &= \arctan \l {k \over k+2} \r \\ \Rightarrow && \sum_{r = 1}^{k+1} \arctan \l {1 \over r^2 + r + 1} \r &= \sum_{r = 1}^k \arctan \l {1 \over r^2 + r + 1} \r+ \arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \l {k \over k+2} \r+\arctan \left ( \frac{1}{(k+1)^2+(k+1)+1} \right) \\ &&&= \arctan \left ( \frac{{k \over k+2}+\frac{1}{(k+1)^2+(k+1)+1} }{1-\frac{k}{k+2}\frac{1}{(k+1)^2+(k+1)+1} } \right) \\ &&&= \arctan \left ( \frac{k((k+1)^2+k+1+k)+(k+2) }{(k+2)((k+1)^2+(k+1)+1)-k} \right) \\ &&&= \arctan \left ( \frac{k^3+3k^2+4k+2 }{k^3+5k^2+8k+6} \right) \\ &&&= \arctan \left ( \frac{(k+1)(k^2+2k+2) }{(k+3)(k^2+2k+2)} \right) \\ &&&= \arctan \left ( \frac{k+1 }{(k+1)+2} \right) \\ \end{align*} Therefore it is true for \(n = k+1\), therefore it is true for all \(n \geq 1\) by the principle of mathematical induction. \begin{align*} && S &= \lim_{n \to \infty} \sum_{r = 1}^n \arctan \l {1 \over r^2 + r + 1} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{n}{n+2} \r \\ &&&= \lim_{n \to \infty} \arctan \l \frac{1}{1+2/n} \r \\ &&&=\arctan\l \lim_{n \to \infty} \frac{1}{1+2/n} \r \\ &&&= \frac{\pi}{4} \end{align*} \begin{align*} && \sum_{r = 1}^\infty \arctan \l {1 \over r^2 - r + 1} \r &= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ (r+1)^2 - (r+1) + 1} \right) \\ &&&= \sum_{r = 0}^\infty \arctan \left( \frac{1}{ r^2+r+1} \right) \\ &&&= \arctan \l \frac{1}{0^2+0+1} \r + \frac{\pi}{4} \\ &&&= \frac{\pi}{2} \end{align*}
2001 Paper 2 Q2
Sketch the graph of the function \([x/N]\), for \(0 < x < 2N\), where the notation \([y]\) means the integer part of \(y\). (Thus \([2.9] = 2\), \ \([4]=4\).)
- Prove that \[ \sum_{k=1}^{2N} (-1)^{[k/N]} k = 2N-N^2. \]
- Let \[ S_N = \sum_{k=1}^{2N} (-1)^{[k/N]} 2^{-k}. \] Find \(S_N\) in terms of \(N\) and determine the limit of \(S_N\) as \(N\to\infty\).
2001 Paper 3 Q12
A bag contains \(b\) black balls and \(w\) white balls. Balls are drawn at random from the bag and when a white ball is drawn it is put aside.
- If the black balls drawn are also put aside, find an expression for the expected number of black balls that have been drawn when the last white ball is removed.
- If instead the black balls drawn are put back into the bag, prove that the expected number of times a black ball has been drawn when the first white ball is removed is \(b/w\,\). Hence write down, in the form of a sum, an expression for the expected number of times a black ball has been drawn when the last white ball is removed.
1999 Paper 2 Q4
By considering the expansions in powers of \(x\) of both sides of the identity $$ {(1+x)^n}{(1+x)^n}\equiv{(1+x)^{2n}}, $$ show that $$ \sum_{s=0}^n {n\choose s}^2 = {2n\choose n}, $$ where \(\displaystyle {n\choose s}= \frac{n!}{s!\,(n-s)!}\). By considering similar identities, or otherwise, show also that:
- if \(n\) is an even integer, then \(\displaystyle \sum_{s=0}^n {{(-1)}^s}{n \choose s}^2= (-1)^{n/2}{n \choose n/2};\)
- \(\displaystyle \sum\limits_{t=1}^ n 2t { n \choose t}^2 = n {2n\choose n} .\)
Solution: To obtain the coefficient of \(x^n\) on the RHS we clearly have \(\displaystyle \binom{2n}n\). To obtain the coefficient of \(x^n\) on the LHS we can obtain \(x^s\) from the first bracket and \(x^{n-s}\) from the second bracket, ie \(\displaystyle \sum_{s=0}^n \binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n \binom{n}{s}\binom{n}{s} = \sum_{s=0}^n \binom{n}{s}^2\)
- Consider \((1-x)^n(1+x)^n = (1-x^2)^n\), then the coefficient of \(x^n\) (if \(n\) is even) is for the RHS \(\displaystyle (-1)^{n/2} \binom{n}{n/2}\). For the LHS, we can obtain \(x^n\) via \(x^s\) and \(x^{n-s}\) which is \(\displaystyle \sum_{s=0}^n (-1)^s\binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n (-1)^s\binom{n}{s}^2\)
- Notice that \begin{align*} && \sum_{t=1}^ n 2t { n \choose t}^2 &= n {2n\choose n} \\ \Leftrightarrow && \sum_{t=1}^ n 2t \frac{n}{t} { n-1 \choose t-1}\binom{n}{t} &= n \frac{2n}{n}{2n-1\choose n-1} \\ \Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{t} &= {2n-1\choose n-1} \\ \Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{n-t} &= {2n-1\choose n-1} \\ \end{align*} but this is exactly what we would obtain by considering the coefficient of \(x^{n-1}\) in \((1+x)^{n-1}(1+x)^n \equiv (1+x)^{2n-1}\)
1999 Paper 2 Q8
Prove that $$ \sum_{k=0}^n \sin k\theta = \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta}\;. \tag{*}$$
- Deduce that, when \(n\) is large, \[ \sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) \approx \frac{2n}\pi\;. \]
- By differentiating \((*)\) with respect to \(\theta\), or otherwise, show that, when \(n\) is large, \[ \sum_{k=0}^n k \sin^2 \left(\frac{k\pi}{2n}\right) \approx \left(\frac{1}4 +\frac{1}{\pi^2} \right)n^2\;. \]
Solution: \begin{align*} && \sum_{k=0}^n \sin k\theta &= \textrm{Im} \left ( \sum_{k=0}^n e^{i k \theta}\right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i \theta}-1} \right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+\tfrac12)\theta}-e^{-i\theta/2}}{e^{i \theta/2}-e^{-i \theta/2}} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/2i}{(e^{i \theta/2}-e^{-i \theta/2})/2i} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/i}{2\sin \tfrac12 \theta} \right)\\ &&&= \frac{\cos \tfrac12 \theta - \cos(n+ \tfrac12)\theta}{2\sin \tfrac12 \theta} \end{align*}
- When \(n\) is large we have \begin{align*} &&\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) &= \frac{\cos \frac{\pi}{2n} - \cos \frac{(2n + 1)\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&= \frac{\cos \frac{\pi}{2n} +\cos \frac{\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&\approx \frac{1- \frac{\pi^2}{4n^2}}{\frac{\pi}{2n}} \\ &&&= \frac{2n}{\pi} - \frac{\pi}{2n} \\ &&&\approx \frac{2n}{\pi} \end{align*}
- \(\,\) \begin{align*} && \sum_{k=0}^n \sin k\theta &= \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta} \\ \frac{\d }{\d \theta}: && \sum_{k=0}^n k\cos k\theta &= \frac { (-\tfrac12\sin\tfrac12\theta +(n+\tfrac12) \sin (n+ \tfrac12) \theta)2 \sin \tfrac12 \theta - (\cos \tfrac12\theta - \cos (n+ \tfrac12) \theta) \cos \tfrac12 \theta} {4\sin^2 \tfrac12\theta} \\ &&&= \frac{-\sin^2 \tfrac12 \theta+(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta-\cos^2\tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta-1}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \\ \Rightarrow && \sum_{k=0}^n k\left ( 1-2\sin^2 \left ( \frac{k\theta}{2} \right) \right) &= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \Rightarrow && \sum_{k=0}^n k\sin^2 \left ( \frac{k\theta}{2} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{8 \sin^2 \tfrac12 \theta} \\ \theta = \frac{\pi}{n}: && \sum_{k=0}^n k\sin^2 \left ( \frac{k\pi}{2n} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\frac{\pi}{n}\sin \tfrac12 \frac{\pi}{n}+\cos n \frac{\pi}{n}-1}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n(n+1)}{4} - \frac{-2n \sin^2 \frac{\pi}{2n}-2}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n^2+n}{4} + \frac{n}{4} + \frac{1}{4\sin^2\frac{\pi}{2n}} \\ &&&\approx \frac{n^2}4 + \frac{n}{2}+ \frac{n^2}{\pi^2} \\ &&&= \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 + \frac{n}{4} \\ &&&\approx \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 \end{align*}
1999 Paper 3 Q3
Justify, by means of a sketch, the formula $$ \lim_{n\rightarrow\infty}\left\{{1\over n}\sum_{m=1}^n \f(1+m/n)\right\} = \int_1^2 \f(x)\,\d x \,. $$ Show that $$ \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} = \ln 2 \,. $$ Evaluate $$ \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\}\,. $$
Solution:
1999 Paper 3 Q12
In the game of endless cricket the scores \(X\) and \(Y\) of the two sides are such that \[ \P (X=j,\ Y=k)=\e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!},\] for some positive constant \(\lambda\), where \(j,k = 0\), \(1\), \(2\), \(\ldots\).
- Find \(\P(X+Y=n)\) for each \(n>0\).
- Show that \(2\lambda \e^{2\lambda-1}=1\).
- Show that \(2x \e^{2x-1}\) is an increasing function of \(x\) for \(x>0\) and deduce that the equation in (ii) has at most one solution and hence determine \(\lambda\).
- Calculate the expectation \(\E(2^{X+Y})\).
Solution:
- \begin{align*} && \mathbb{P}(X+Y = n) &= \sum_{i = 0}^n \mathbb{P}(X = i, Y = n-i) \\ &&&= \sum_{i = 0}^n e^{-1} \frac{n \lambda^n}{i! (n-i)!} \\ &&&=e^{-1} n \lambda^n \sum_{i = 0}^n\frac{1}{i! (n-i)!} \\ &&&=\frac{e^{-1} n}{n!} \lambda^n \sum_{i = 0}^n\frac{n!}{i! (n-i)!} \\ &&&= \frac{n\lambda^n}{e n!} 2^n \\ &&&= \frac{n (2 \lambda)^n}{e \cdot n!} \end{align*}
- \begin{align*} && 1 &= \sum_{n = 0}^{\infty} \mathbb{P}(X+Y =n ) \\ &&&= \sum_{n = 0}^{\infty}\frac{n (2 \lambda)^n}{e \cdot n!} \\ &&&= \sum_{n = 1}^\infty \frac{ (2 \lambda)^n}{e \cdot (n-1)!} \\ &&&= \frac{2 \lambda}{e}\sum_{n = 0}^\infty \frac{ (2 \lambda)^n}{n!} \\ &&&= \frac{2 \lambda}{e} e^{2\lambda} \\ &&&= 2 \lambda e^{2\lambda - 1} \end{align*} \\
- Consider \(f(x) = 2xe^{2x-1}\), then \begin{align*} && f'(x) &= 2e^{2x-1} + 2xe^{2x-1} \cdot 2 \\ &&&= e^{2x-1} (2 + 4x) > 0 \end{align*} Therefore \(f(x)\) is an increasing function of \(x\), which means \(f(x) = 1\) has at most one solution for \(\lambda\). Therefore \(\lambda = \frac12\)
- \begin{align*} \mathbb{E}(2^{X+Y}) &= \sum_{n = 0}^\infty \mathbb{P}(X+Y = n) 2^n \\ &= \sum_{n = 1}^\infty \frac{1}{e(n-1)!} 2^{n} \\ &= \frac{2}{e} \sum_{n=0}^\infty \frac{2^n}{n!} \\ &= \frac{2}{e} e^2 \\ &= 2e \end{align*}
1999 Paper 3 Q14
In the basic version of Horizons (H1) the player has a maximum of \(n\) turns, where \(n \ge 1\). At each turn, she has a probability \(p\) of success, where \(0 < p < 1\). If her first success is at the \(r\)th turn, where \(1 \le r \le n\), she collects \(r\) pounds and then withdraws from the game. Otherwise, her winnings are nil. Show that in H1, her expected winnings are $$ p^{-1}\left[1+nq^{n+1}-(n+1)q^n\right]\quad\hbox{pounds}, $$ where \(q=1-p\). The rules of H2 are the same as those of H1, except that \(n\) is randomly selected from a Poisson distribution with parameter \(\lambda\). If \(n=0\) her winnings are nil. Otherwise she plays H1 with the selected \(n\). Show that in H2, her expected winnings are $$ {1 \over p}{\left(1-{\e^{-{\lambda}p}}\right)} -{{\lambda}q}{\e^{-{\lambda}p}} \quad\hbox{pounds}. $$
Solution: \begin{align*} && \E[H1] &= \sum_{r=1}^n r \cdot \mathbb{P}(\text{first success on }r\text{th turn}) \\ &&&= \sum_{r=1}^n r \cdot q^{r-1}p \\ &&&= p\sum_{r=1}^n r q^{r-1} \\ \\ && \frac{1-x^{n+1}}{1-x} &= \sum_{r=0}^n x^r \\ \Rightarrow && \sum_{r=1}^n r x^{r-1} &= \frac{-(n+1)x^n(1-x) +(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} \\ \\ && \E[H1] &= p\sum_{r=1}^n r q^{r-1} \\ &&&= p\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} \\ &&&= p^{-1}(1-(n+1)q^{n} + nq^{n+1}) \end{align*} Not that if \(n =0\) , the formula for \(\E[H1] = 0\). So \begin{align*} && \E[H2] &= \E[\E[H1|n=N]] \\ &&&= p^{-1}\E \left [ 1-(N+1)q^{N} + Nq^{N+1}\right] \\ &&&= p^{-1}\E \left [ 1-((1-q)N+1)q^{N} \right] \\ &&&= p^{-1}\left (1 - p\E[Nq^N] - G_{Po(\lambda)}(q) \right) \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - \E[Nq^N] \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - q\lambda e^{-\lambda(1-q)} \\ &&&= p^{-1}(1-e^{-\lambda p}) - q\lambda e^{-\lambda p} \end{align*}
1998 Paper 1 Q7
My bank pays \(\rho\%\) interest at the end of each year. I start with nothing in my account. Then for \(m\) years I deposit \(\pounds a\) in my account at the beginning of each year. After the end of the \(m\)th year, I neither deposit nor withdraw for \(l\) years. Show that the total amount in my account at the end of this period is \[\pounds a\frac{r^{l+1}(r^{m}-1)}{r-1}\] where \(r=1+{\displaystyle \frac{\rho}{100}}\). At the beginning of each of the \(n\) years following this period I withdraw \(\pounds b\) and this leaves my account empty after the \(n\)th withdrawal. Find an expression for \(a/b\) in terms of \(r\), \(l\), \(m\) and \(n\).
Solution: Rather than putting the deposits in the same account, imagine they are all put in separate accounts. Then for example, the first \(\pounds a\) will go on to become \(\pounds r^m \cdot r^l a\) from \(m\) years of compound interest as more money is deposited, followed by \(l\) years where no money is deposited. Therefore the total amount at the end is: \begin{align*} && S &= r^{m}r^l a + r^{m-1}r^l a + \cdots + r r^l a \\ &&&= r^l a(r^m + \cdots + r) \\ &&&= ar^l\frac{r^{m+1}-r}{r-1} \\ &&&= a \frac{r^{l+1}(r^m-1)}{r-1} \end{align*} Rather than withdrawing \(b\) each time, imagine that in the \(n\)th year we withdraw each \(b\) with the appropriate additional interest, ie \begin{align*} && \underbrace{a \frac{r^{l+1}(r^m-1)}{r-1}}_{\text{amount before \(n\) years}} \underbrace{r^{n-1}}_{\text{accounting for interest}} &= b r^{n-1} + br^{n-2} + \cdots + b \\ &&&= b \frac{r^n-1}{r-1} \\ \Rightarrow && \frac{a}{b} &= \frac{r^n-1}{r^{l+n}(r^m-1)} \end{align*}
1998 Paper 2 Q3
Show that the sum \(S_N\) of the first \(N\) terms of the series $$\frac{1}{1\cdot2\cdot3}+\frac{3}{\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+\cdots +\frac{2n-1}{n(n+1)(n+2)}+\cdots$$ is $${1\over2}\left({3\over2}+{1\over N+1}-{5\over N+2}\right).$$ What is the limit of \(S_N\) as \(N\to\infty\)? The numbers \(a_n\) are such that $$\frac{a_n}{a_{n-1}}=\frac{(n-1)(2n-1)}{(n+2)(2n-3)}.$$ Find an expression for \(a_n/a_1\) and hence, or otherwise, evaluate \(\sum\limits_{n=1}^\infty a_n\) when \(\displaystyle a_1=\frac{2}{9}\;\).
Solution: First notice by partial fractions: \begin{align*} \frac{2n-1}{n(n+1)(n+2)} &= \frac{-1/2}{n} + \frac{3}{n+1} + \frac{-5/2}{n+2} \\ &= \frac{-1}{2n} + \frac{3}{n+1} - \frac{5}{2(n+2)} \end{align*} And therefore: \begin{align*} \sum_{n = 1}^N \frac{2n-1}{n(n+1)(n+2)} &= -\frac12 \sum_{n=1}^N \frac1n +3\sum_{n=1}^N \frac1{n+1} -\frac52 \sum_{n=1}^N \frac1{n+2} \\ &= -\frac12-\frac14 + \frac{3}{2}+ \sum_{n=3}^N (3-\frac12 -\frac52)\frac1n + \frac{3}{N+1} - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} \\ &= \frac12 \l \frac32+\frac1{N+1}-\frac{5}{N+2} \r \end{align*} As \(N \to \infty, S_N \to \frac{3}{4}\). \begin{align*} && \frac{a_n}{a_{n-1}}&=\frac{(n-1)(2n-1)}{(n+2)(2n-3)} \\ \Rightarrow && \frac{a_n}{a_1} &= \frac{a_n}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1} \\ &&&= \frac{(n-1)(2n-1)}{(n+2)(2n-3)} \cdot \frac{(n-2)(2n-3)}{(n+1)(2n-5)} \cdots \frac{(1)(3)}{(4)(1)} \\ &&&= \frac{(2n-1)3\cdot 2\cdot 1}{(n+2)(n+1)n} \\ &&& = \frac{6(2n-1)}{n(n+1)(n+2)} \end{align*} Therefore \(a_n = \frac{4}{3} \frac{2n-1}{n(n+1)(n+2)}\) and so our sequence is \(\frac43\) the earlier sum, ie \(1\)
1997 Paper 2 Q1
Find the sum of those numbers between 1000 and 6000 every one of whose digits is one of the numbers \(0,\,2,\,5\) or \(7\), giving your answer as a product of primes.
Solution: The first digit is \(2\) or \(5\), all the other digits can be any value from \(0,2,5,7\). Therefore we have \begin{align*} S &= 2000 \cdot 4^3+5000 \cdot 4^3 + (200+500+700) \cdot 2 \cdot 4^2 + (20+50+70) \cdot 2 \cdot 4^2 + (2+5+7) \cdot 2 \cdot 4^2 \\ &= 7 \cdot 4^3 \cdot 2^3 \cdot 5^3 + 14 \cdot 2 \cdot 4^2 \cdot 111 \\ &= 2^{9} \cdot5^3 \cdot 7 + 2^{6} \cdot 3 \cdot 7 \cdot 37 \\ &= 2^6 \cdot 7 \cdot (1000+111) \\ &= 2^6 \cdot 7 \cdot 11 \cdot 101 \end{align*} Alternatively, consider adding the first and last terms, and second and second and last terms, etc we obtain \(7777\). There are \(2 \cdot 4^3\) terms so \(7777 \cdot 4^3 = 2^6 \cdot 7 \cdot 11 \cdot 101\)
1997 Paper 3 Q3
By considering the solutions of the equation \(z^n-1=0\), or otherwise, show that \[(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})=1+z+z^2+\dots+z^{n-1},\] where \(z\) is any complex number and \(\omega={\rm e}^{2\pi i/n}\). Let \(A_1,A_2,A_3,\dots,A_n\) be points equally spaced around a circle of radius \(r\) centred at \(O\) (so that they are the vertices of a regular \(n\)-sided polygon). Show that \[\overrightarrow{OA_1}+\overrightarrow{OA_2}+\overrightarrow{OA_3} +\dots+\overrightarrow{OA_n}=\mathbf0.\] Deduce, or prove otherwise, that \[\sum_{k=1}^n|A_1A_k|^2=2r^2n.\]