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2024 Paper 2 Q6
D: 1500.0 B: 1500.0
generalised binomial theorem proof by induction recurrence relation binomial coefficients power series cauchy product

In this question, you need not consider issues of convergence.

  1. The sequence \(T_n\), for \(n = 0, 1, 2, \ldots\), is defined by \(T_0 = 1\) and, for \(n \geqslant 1\), by \[ T_n = \frac{2n-1}{2n}\,T_{n-1}. \] Prove by induction that \[ T_n = \frac{1}{2^{2n}}\binom{2n}{n}, \] for \(n = 0, 1, 2, \ldots\). [Note that \(\dbinom{0}{0} = 1\).]
  2. Show that in the binomial series for \((1-x)^{-\frac{1}{2}}\), \[ (1-x)^{-\frac{1}{2}} = \sum_{r=0}^{\infty} a_r x^r, \] successive coefficients are related by \[ a_r = \frac{2r-1}{2r}\,a_{r-1} \] for \(r = 1, 2, \ldots\)\,. Hence prove that \(a_r = T_r\) for all \(r = 0, 1, 2, \ldots\)\,.
  3. Let \(b_r\) be the coefficient of \(x^r\) in the binomial series for \((1-x)^{-\frac{3}{2}}\), so that \[ (1-x)^{-\frac{3}{2}} = \sum_{r=0}^{\infty} b_r x^r. \] By considering \(\dfrac{b_r}{a_r}\), find an expression involving a binomial coefficient for \(b_r\), for \(r = 0, 1, 2, \ldots\)\,.
  4. By considering the product of the binomial series for \((1-x)^{-\frac{1}{2}}\) and \((1-x)^{-1}\), prove that \[ \frac{(2n+1)}{2^{2n}}\binom{2n}{n} = \sum_{r=0}^{n} \frac{1}{2^{2r}}\binom{2r}{r}, \] for \(n = 1, 2, \ldots\)\,.

Solution:

  1. Claim: \(\displaystyle T_n = \frac{1}{2^{2n}}\binom{2n}{n}\) Proof: (By Induction) Base case: \(n=0\). Note that \(T_0 = 1\) and \(\frac{1}{2^0}\binom{0}{0} = 1\) so the base case is true. Assume true for some \(n=k\), ie \(T_k = \frac{1}{2^{2k}} \binom{2k}{k}\) so \begin{align*} && T_{k+1} &= \frac{2(k+1)-1}{2(k+1)} \frac{1}{2^{2k}} \binom{2k}{k} \\ &&&= \frac{2k+1}{k+1} \frac{1}{2^{2k+1}} \frac{(2k)!}{k!k!} \\ &&&= \frac{2(k+1)(2k+1)}{(k+1)(k+1)} \frac{1}{2^{2(k+1)}} \frac{(2k)!}{k!k!} \\ &&&= \frac{1}{2^{2(k+1)}} \frac{(2k+2)!}{(k+1)!(k+1)!} \\ &&&= \frac{1}{2^{2(k+1)}} \binom{2(k+1)}{k+1} \end{align*} and therefore it's true for all \(n\).
  2. Notice that \((1-x)^{-\frac12} = 1 + (-\tfrac12)(-x) + \frac{(-\frac12)(-\frac32)}{2!}(-x)^2+\cdots\) in particular \(a_r = \frac{(-\frac12 - r)}{r}(-1)a_{r-1} = \frac{2r-1}{2r}a_{r-1}\). Since \(a_0 = 1\) we have \(a_r = T_r\) for all \(r\).
  3. Notice that \begin{align*} && (1-x)^{-\frac32} &= \sum_{r=0}^\infty b_r x^r \\ &&&= \sum_{r=0}^\infty \frac{(-\frac32)\cdot(-\frac32-1)\cdots (-\frac32-(r-1))}{r!}(-x)^r \\ &&&= \sum_{r=0}^\infty \frac{(-\frac12-1)\cdot(-\frac12-2)\cdots (-\frac12-r)}{r!}(-x)^r \\ \end{align*} Therefore \(\frac{b_r}{a_r} = \frac{r+\frac12}{\frac12} = 2r+1\) so \(b_r = \frac{2r+1}{2^{2r}} \binom{2r}{r}\)
  4. Notice that \begin{align*} && (1-x)^{-\frac32} &= (1-x)^{-\frac12}(1-x)^{-1} \\ &&&= (1 + x+ x^2 + \cdots) \sum_{r=0}^{\infty} a_r x^r \\ &&&= \sum_{i=0}^{\infty} \sum_{k=0}^n a_r x^i \end{align*} So we must have \(b_r = \sum_{i=0}^ra_i\) which is the required result

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2016 Paper 1 Q8
D: 1500.0 B: 1530.6
generating functions generalised binomial theorem binomial series recurrence relation power series Fibonacci sequence partial fractions sequences

Given an infinite sequence of numbers \(u_0\), \(u_1\), \(u_2\), \(\ldots\,\), we define the generating function, \(\f\), for the sequence by \[ \f(x) = u_0 + u_1x +u_2 x^2 +u_3 x^3 + \cdots \,. \] Issues of convergence can be ignored in this question.

  1. Using the binomial series, show that the sequence given by \(u_n=n\,\) has generating function \(x(1-x)^{-2}\), and find the sequence that has generating function \(x(1-x)^{-3}\). Hence, or otherwise, find the generating function for the sequence \(u_n =n^2\). You should simplify your answer.
    • \(\bf (a)\) The sequence \(u_0\), \(u_1\), \(u_2\), \(\ldots\,\) is determined by \(u_{n} = ku_{n-1}\) (\(n\ge1\)), where \(k\) is independent of \(n\), and \(u_0=a\). By summing the identity \(u_{n}x^n \equiv ku_{n-1}x^n\), or otherwise, show that the generating function, f, satisfies \[ \f(x) = a + kx \f(x) \] Write down an expression for \(\f(x)\).
    • \(\bf (b)\) The sequence \(u_0, u_1, u_2, \ldots\,\) is determined by \(u_{n} = u_{n-1}+ u_{n-2}\) (\(n\ge2\)) and \(u_0=0\), \(u_1=1\). Obtain the generating function.

Solution:

  1. \(\,\) \begin{align*} && x(1-x)^{-2} &= x \left (1 + (-2)(-x) + \frac{(-2)(-3)}{2!}x^2 + \cdots + \frac{(-2)(-3)\cdots(-2-(k-1))}{k!} (-x)^k + \cdots \right) \\ &&&= x(1 + 2x + 3x^2 + \cdots + \frac{(-2)(-3)\cdots(-(k+1))}{k!}(-1)^k x^k + \cdots ) \\ &&&= x+2x^2 + 3x^3 + \cdots + (k+1)x^{k+1} + \cdots \\ \Rightarrow && u_n &= n \end{align*} \begin{align*} && x(1-x)^{-3} &= x \left (1 + 3x + 6x^2 + \cdots + \frac{(-3)(-4)\cdots(-k-2)}{k!}(-x)^k + \cdots \right) \\ &&&= x \left (1 + 3x + 6x^2 + \cdots + \frac{(k+2)(k+1)}{2}x^k + \cdots \right) \\ &&&= x + 3x^2 + 6x^3 + \cdots + \binom{k+2}{2}x^{k+1} + \cdots \\ && u_n &= \binom{n+1}{2} = \frac{n^2+n}{2} \\ \\ \Rightarrow && 2x(1-x)^{-3} - x(1-x)^{-2} &= (1-x)^{-3}(2x-x(1-x)) \\ &&&= (1-x)^{-3}(x+x^2) \end{align*}
    • \(u_n = ku_{n-1} \Rightarrow u_nx^n = ku_{n-1}x^n\) so \begin{align*} && \sum_{n=1}^\infty u_n x^n &= \sum_{n=1}^\infty k u_{n-1}x^n \\ && \sum_{n=0}^\infty u_n x^n - a &= x\sum_{n=0}^\infty k u_{n}x^n \\ \Rightarrow && f(x)-a &= kx f(x) \\ \Rightarrow && f(x) &= a + kxf(x) \\ \Rightarrow && f(x) &= \frac{a}{1-kx} \end{align*}
    • Suppose \(\displaystyle f(x) = \sum_{n=0}^\infty u_n x^n\) so \begin{align*} && x^n u_n &= x^n u_{n-1} + x^n u_{n-2} \\ \Rightarrow && \sum_{n=2}^\infty x^n u_n &= \sum_{n=2}^\infty x^n u_{n-1} + \sum_{n=2}^\infty x^n u_{n-2} \\ && \sum_{n=0}^\infty x^n u_n - u_0 - u_1 x &= \left ( \sum_{n=0}^\infty x^{n+1} u_{n} -xu_0 \right) + \sum_{n=0}^\infty x^{n+2} u_{n} \\ && f(x) - x &= xf(x) +x^2f(x) \\ \Rightarrow && f(x) &= \frac{x}{1-x-x^2} \end{align*}

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2011 Paper 1 Q6
D: 1500.0 B: 1500.0
generalised binomial theorem negative binomial expansion partial fractions power series coefficient extraction summation of series polynomial division

Use the binomial expansion to show that the coefficient of \(x^r\) in the expansion of \((1-x)^{-3}\) is \(\frac12 (r+1)(r+2)\,\).

  1. Show that the coefficient of \(x^r\) in the expansion of \[ \frac{1-x+2x^2}{(1-x)^3} \] is \(r^2+1\) and hence find the sum of the series \[ 1+\frac22+\frac54+\frac{10}8+\frac{17}{16}+\frac{26}{32}+\frac{37}{64} +\frac{50}{128}+ \cdots \,. \]
  2. Find the sum of the series \[ 1+2+\frac94+2+\frac{25}{16}+\frac{9}{8}+\frac{49}{64} + \cdots \,. \]

Solution: Notice that the coefficient of \(x^r\) is \((-1)^r\frac{(-3) \cdot (-3-1) \cdots (-3-r+1)}{r!} = (-1)^r \frac{(-1)(-2)(-3)(-4) \cdots (-(r+2))}{(-1)(-2)r!} = (-1)^r(-1)^{r+2}\frac{(r+2)!}{2r!} = \frac{(r+2)(r+1)}2\).

  1. The coefficient of \(x^r\) is \begin{align*} && c_r &=\frac{(r+1)(r+2)}{2} - \frac{(r-1+1)(r-1+2)}{2} + 2 \frac{((r-2+1)(r-2+2)}{2} \\ &&&= \frac{r^2+3r+2}{r} - \frac{r^2+r}{2} + \frac{2r^2-2r}{2}\\ &&&= \frac{2r^2+2}{2} = r^2+1 \end{align*} \begin{align*} && S & = 1+\frac22+\frac54+\frac{10}8+\frac{17}{16}+\frac{26}{32}+\frac{37}{64} +\frac{50}{128}+ \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{r^2+1}{2^r} \\ &&&= \frac{1-\tfrac12+2 \cdot \tfrac14}{(1-\tfrac12)^3} \\ &&&= 8 \end{align*}
  2. \(\,\) \begin{align*} && S &= 1+2+\frac94+2+\frac{25}{16}+\frac{9}{8}+\frac{49}{64} + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^r} \\ &&&= 2 \sum_{r=0}^{\infty} \frac{(r+1)^2}{2^{r+1}} \\ &&&= 2 \sum_{r=1}^{\infty} \frac{r^2}{2^{r}} \\ &&&= 2 \left (\sum_{r=0}^{\infty} \frac{r^2+1}{2^{r}} - \sum_{r=0}^{\infty} \frac{1}{2^{r}} \right) \\ &&&= 2 (8 - 1) = 14 \end{align*}

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2009 Paper 3 Q2
D: 1700.0 B: 1484.0
second order differential equation power series series solution recurrence relation comparing coefficients trigonometric Maclaurin series

  1. Let \(\displaystyle y= \sum_{n=0}^\infty a_n x^n\,\), where the coefficients \(a_n\) are independent of \(x\) and are such that this series and all others in this question converge. Show that \[ \displaystyle y'= \sum_{n=1}^\infty na_n x^{n-1}\,, \] and write down a similar expression for \(y''\). Write out explicitly each of the three series as far as the term containing \(a_3\).
  2. It is given that \(y\) satisfies the differential equation \[ xy''-y'+4x^3y =0\,. \] By substituting the series of part (i) into the differential equation and comparing coefficients, show that \(a_1=0\). Show that, for \(n\ge4\), \[ a_n =- \frac{4}{n(n-2)}\, a_{n-4}\,, \] and that, if \(a_0=1\) and \(a_2=0\), then \( y=\cos (x^2)\,\). Find the corresponding result when \(a_0=0\) and \(a_2=1\).

Solution:

  1. Let \(\displaystyle y= \sum_{n=0}^\infty a_n x^n\,\) then \begin{align*} y' &= \frac{\d}{\d x} \l \sum_{n=0}^\infty a_n x^n \r \\ &= \sum_{n=0}^\infty \frac{\d}{\d x} \l a_n x^n \r \\ &= \sum_{n=0}^\infty n a_n x^{n-1} \\ &= \sum_{n=1}^\infty n a_n x^{n-1} \\ \\ y'' &= \frac{\d}{\d x} \l\sum_{n=1}^\infty n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty \frac{\d}{\d x} \l n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty n(n-1) a_n x^{n-2} \\ &= \sum_{n=2}^\infty n(n-1) a_n x^{n-2} \\ \\ y &= a_0 + a_1 x+ a_2x^2 + a_3x^3 + \cdots \\ y'&= a_1 + 2a_2x+3a_3x^2 + \cdots \\ y'' &= 2a_2 + 6a_3x + \cdots \end{align*}
  2. \begin{align*} && 0 &= xy''-y'+4x^3y \\ &&&= x\sum_{n=2}^\infty n(n-1) a_n x^{n-2} - \sum_{n=1}^\infty n a_n x^{n-1} + 4x^3 \sum_{n=0}^\infty a_n x^n \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=0}^\infty 4a_n x^{n+3} \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=4}^\infty 4a_{n-4} x^{n-1} \\ &&&= \sum_{n=4}^{\infty} \l n(n-1) a_n- n a_n +4a_{n-4} \r x^{n-1} + 2a_2x + 6a_3x^2-a_1-2a_2x-3a_3x^2 \\ &&&= \sum_{n=4}^{\infty} \l n(n-2) a_n +4a_{n-4} \r x^{n-1}+ 3a_3x^2-a_1 \\ \end{align*} Therefore since all coefficients are \(0\), \(a_1 = 0\), \(a_3 = 0\) and \(\displaystyle a_n = -\frac{4}{n(n-2)}a_{n-4}\). If \(a_0 = 1, a_2 = 0\), and since \(a_1 = 0, a_3 = 0\) the only values which will take non-zero value are \(a_{4k}\). We can compute these values as: \(a_{4k} = -\frac{4}{(4k)(4k-2)} a_{4k-4} = \frac{1}{2k(2k-1)}a_{4k-r}\) so \(a_{4k} = \frac{(-1)^k}{(2k)!}\), which are precisely the coefficients in the expansion \(\cos x^2\). If \(a_0 = 0, a_2 = 1\) then since \(a_1 = 0, a_3 = 0\) the only values which take non-zero values are \(a_{4k+2}\) we can compute these values as: \(a_{4k+2} = -\frac{4}{(4k+2)(4k)}a_{4k-2} = -\frac{1}{(2k+1)2k}a_{4k-2}\) so we can see that \(a_{4k+2}= \frac{(-1)^k}{(2k+1)!}\) precisely the coefficients of \(\sin x^2\)

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2008 Paper 2 Q2
D: 1600.0 B: 1498.5
partial fractions series expansion binomial expansion geometric series power series decimal approximation rational function

Let \(a_n\) be the coefficient of \(x^n\) in the series expansion, in ascending powers of \(x\), of \[\displaystyle \frac{1+x}{(1-x)^2(1+x^2)} \,, \] where \(\vert x \vert <1\,\). Show, using partial fractions, that either \(a_n =n+1\) or \(a_n = n+2\) according to the value of \(n\). Hence find a decimal approximation, to nine significant figures, for the fraction \( \displaystyle \frac{11\,000}{8181}\). \newline [You are not required to justify the accuracy of your approximation.]

Solution: \begin{align*} && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\ \Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\ \Rightarrow && 2 &= 2B \tag{\(x = 1\)} \\ \Rightarrow && 1 &= B \\ \Rightarrow && 1 &= A+B+D \tag{\(x = 0\)}\\ \Rightarrow && A &= -D \\ \Rightarrow && 0 &= 4A+2B-4C+4D \tag{\(x = -1\)}\\ \Rightarrow && C &= \frac12\\ \Rightarrow && 3 &= -5A+5B+2C+D \tag{\(x=2\)} \\ \Rightarrow && 3 &= -6A+6 \\ \Rightarrow && A,D &=-\frac12,\frac12 \\ \Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\ &&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k} \end{align*} Therefore the coefficient of \(x^n\) is \(n+1\) or \(n+2\) depending on whether the coefficients from the final series add constructively \(n \equiv 1, 2 \pmod{4}\) or destructively. \begin{align*} \frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\ &= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\ &= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\ & \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\ &= 1.34457890 + \frac{12}{10^{10}} + \cdots \end{align*} \begin{align*} && \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\ && &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\ \Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\ &&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right ) \end{align*} Therefore for this will be less than \(10^{-9}\), when \(m = 11\), so our approximation is valid to 9sf

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2008 Paper 3 Q8
D: 1700.0 B: 1500.0
sequences recurrence relation power series geometric series partial fractions sum to infinity finite sum generating functions

  1. The coefficients in the series \[ S= \tfrac13 x + \tfrac 16 x^2 + \tfrac1{12} x^3 + \cdots + a_rx^r + \cdots \] satisfy a recurrence relation of the form \(a_{r+1} + p a_r =0\). Write down the value of \(p\). By considering \((1+px)S\), find an expression for the sum to infinity of \(S\) (assuming that it exists). Find also an expression for the sum of the first \(n+1\) terms of \(S\).
  2. The coefficients in the series \[ T=2 + 8x +18x^2+37 x^3 +\cdots + a_rx^r + \cdots \] satisfy a recurrence relation of the form \(a_{r+2}+pa_{r+1} +qa_r=0\). Find an expression for the sum to infinity of \(T\) (assuming that it exists). By expressing \(T\) in partial fractions, or otherwise, find an expression for the sum of the first \(n+1\) terms of \(T\).

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2007 Paper 3 Q2
D: 1700.0 B: 1516.0
generalised binomial theorem double factorial differentiation power series summation binomial expansion factorial identity convergence

  1. Show that \(1.3.5.7. \;\ldots \;.(2n-1)=\dfrac {(2n)!}{2^n n!}\;\) and that, for $\vert x \vert < \frac14$, \[ \frac{1}{\sqrt{1-4x\;}\;} =1+\sum_{n=1}^\infty \frac {(2n)!}{(n!)^2} \, x^n \,. \]
  2. By differentiating the above result, deduce that \[ \sum _{n=1}^\infty \frac{(2n)!}{n!\,(n-1)!} \left(\frac6{25}\right)^{\!\!n} = 60 \,. \]
  3. Show that \[ \sum _{n=1}^\infty \frac{2^{n+1}(2n)!}{3^{2n}(n+1)!\,n!} = 1 \,. \]

Solution:

  1. Notice that \(1 \cdot 3 \cdot 5 \cdot 7 \cdot (2n-1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots \cdots 2n}{2 \cdot 4 \cdot 6 \cdots 2n} = \frac{(2n)!}{2^n \cdot n!}\) as required. When \(|4x| < 1\) or \(|x|<\frac14\) we can apply the generalised binomial theorem to see that: \begin{align*} \frac{1}{\sqrt{1-4x}} &= (1-4x)^{-\frac12} \\ &= 1+\sum_{n=1}^\infty \frac{-\frac12 \cdot \left ( -\frac32\right)\cdots \left ( -\frac{2n-1}2\right)}{n!} (-4x)^n \\ &= 1+\sum_{n=1}^{\infty} (-1)^n\frac{(2n)!}{(n!)^2 2^{2n}} (-4)^n x^n \\ &= 1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } x^n \\ \end{align*}
  2. Differentiating we obtain \begin{align*} && 2(1-4x)^{-\frac32} &= \sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} x^{n-1} \\ \Rightarrow &&\sum_{n=1}^\infty \frac{(2n)!}{n!(n-1)!} \left (\frac{6}{25} \right)^{n} &= \frac{6}{25} \cdot 2\left(1- 4 \frac{6}{25}\right)^{-\frac32} \\ &&&= \frac{12}{25} \left (\frac{1}{25} \right)^{-\frac32} \\ &&&= \frac{12}{25} \cdot 125 = 60 \end{align*}
  3. By integrating, we obtain \begin{align*} && \int_{t=0}^{t=x} \frac{1}{\sqrt{1-4t}} \d t &= \int_{t=0}^{t=x} \left (1+\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2 } t^n \right) \d t \\ \Rightarrow && \left [ -\frac12 \sqrt{1-4t}\right]_0^x &= x + \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\ \Rightarrow && \frac12 - \frac12\sqrt{1-4x} - x &= \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } x^{n+1} \\ \\ \Rightarrow && 9\sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)! } \left ( \frac{2}{9}\right)^{n+1} &=9 \cdot\left( \frac12 - \frac12 \sqrt{1-4\cdot \frac{2}{9}} - \frac29\right )\\ &&&= 9 \cdot \left (\frac12 - \frac{1}{6} - \frac{2}{9} \right) \\ &&&= 1 \end{align*}

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2006 Paper 3 Q3
D: 1700.0 B: 1500.0
Taylor series trigonometric power series cot tan cosec trigonometric identities series coefficients

  1. Let \[ \tan x = \sum\limits_{n=0}^\infty a_n x^n \text{ and } \cot x = \dfrac 1 x +\sum\limits_{n=0}^\infty b_nx^n \] for \(0< x < \frac12\pi\,\). Explain why \(a_n=0\) for even \(n\). Prove the identity \[ \cot x - \tan x \equiv 2 \cot 2x\, \] and show that \[a_{n} = (1-2^{n+1})b_n\,.\]
  2. Let $ \displaystyle {\rm cosec}\, x = \frac1x +\sum\limits _{n=0}^\infty c_n x^n\,$ for \(0< x < \frac12\pi\,\). By considering \(\cot x + \tan x\), or otherwise, show that \[ c_n = (2^{-n} -1)b_n \,. \]
  3. Show that \[ \left(1+x{ \sum\limits_{n=0}^\infty} b_n x^n \right)^2 +x^2 = \left(1+x{ \sum\limits_{n=0} ^\infty} c_n x^n \right)^2\,. \] Deduce from this and the previous results that \(a_1=1\), and find \(a_3\).

Solution:

  1. Since \(\tan (-x) = -\tan x\), \(\tan\) is an odd function, and in particular all it's even coefficients are zero. \begin{align*} && 2 \cot 2x &\equiv \frac{2 cos 2x}{\sin 2 x} \\ &&&\equiv \frac{2(\cos^2 x- \sin^2 x)}{2 \sin x \cos x} \\ &&&\equiv \frac{\cos x}{\sin x} - \frac{\sin x}{ \cos x} \\ &&&\equiv \cot x - \tan x \end{align*} Therefore \begin{align*} && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} - \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2 \left (\underbrace{\frac{1}{2x} + \sum_{n=0}^\infty b_n(2x)^n}_{\cot 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty a_n x^n &= \sum_{n=0}^\infty b_nx^n - 2\sum_{n=0}^\infty b_n(2x)^n \\ &&&= \sum_{n=0}^{\infty}b_n(1-2^{n+1})x^n \\ [x^n]: && a_n &= (1-2^{n+1})b_n \end{align*}
  2. \(\,\) \begin{align*} && \cot x + \tan x &= \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \\ &&&= \frac{1}{\sin x \cos x} \\ &&&=2\cosec 2x \\ \\ \Rightarrow && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} + \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2\left (\underbrace{ \frac1{2x} +\sum\limits _{n=0}^\infty c_n (2x)^n}_{\cosec 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty 2^{n+1}c_n x^n &= \sum_{n=0}^{\infty}(a_n+b_n)x^n \\ &&&= \sum_{n=0}^{\infty}\left((1-2^{n+1})b_n+ b_n\right)x^n \\ &&&= \sum_{n=0}^{\infty}\left(2-2^{n+1}\right)b_nx^n \\ [x^n]: && c_n &= (2^{-n}-1)b_n \end{align*}
  3. \(\,\) \begin{align*} && \cot^2 x + 1 &= \cosec^2 x \\ \Rightarrow && x^2 \cot^2 x + x^2 &= x^2 \cosec^2 x \\ \Rightarrow && x^2 \left ( \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} \right)^2 + x^2 &= x^2 \left (\underbrace{ \frac1{x} +\sum\limits _{n=0}^\infty c_n x^n}_{\cosec x} \right)^2 \\ \Rightarrow && \left ( 1 + x\sum_{n=0}^\infty b_nx^{n} \right)^2 + x^2 &= \left ( 1 +x\sum\limits _{n=0}^\infty c_n x^{n} \right)^2 \\ \\ \Rightarrow && \left ( 1 + x(b_1x + b_3 x^3 + \cdots) \right)^2 + x^2 &= \left ( 1 + x(c_1x + c_3 x^3 + \cdots) \right)^2 \\ \Rightarrow && 1 + (1+2b_1)x^2+(2b_3+b_1^2)x^4 + \cdots &= 1 + 2c_1x^2 + (2c_3+c_1^2)x^4 + \cdots \\ \Rightarrow && 1 + 2b_1 &= 2(2^{-1}-1)b_1 \\ \Rightarrow && b_1 &= -\frac13 \\ \Rightarrow && a_1 &= (1-2^{2})(-\tfrac13) = 1 \\ && c_1 &= \frac16\\ \Rightarrow && 2b_3+\frac19&= 2c_3+\frac1{36} \\ \Rightarrow && 2b_3 -2(2^{-3}-1)b_3 &= -\frac{1}{12} \\ \Rightarrow && \frac{15}{4}b_3 &= -\frac{1}{12} \\ \Rightarrow && b_3 &= -\frac{1}{45} \\ \Rightarrow && a_3 &= -(1-2^4)\frac{1}{45} = \frac13 \end{align*}

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2005 Paper 2 Q6
D: 1600.0 B: 1500.0
generalised binomial theorem power series binomial expansion summation negative exponent differentiation of series infinite series evaluation

  1. Write down the general term in the expansion in powers of \(x\) of \((1-x)^{-1}\), \((1-x)^{-2}\) and \((1-x)^{-3}\), where \(|x| <1\). Evaluate \(\displaystyle \sum_{n=1}^\infty n 2^{-n}\) and \(\displaystyle \sum_{n=1}^\infty n^22^{-n}\).
  2. Show that $\displaystyle (1-x)^{-\frac12} = \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \frac{x^n}{2^{2n}}\( , for \)|x|<1$. Evaluate \(\displaystyle \sum_{n=0}^\infty \frac{(2n)!} {(n!)^2 2^{2n}3^{n}} \) and \(\displaystyle \sum_{n=1}^\infty \frac{n(2n)!} {(n!)^2 2^{2n}3^{n}}\).

Solution:

  1. \(\displaystyle (1-x)^{-1} = \sum_{n=0}^\infty x^n\), \(\displaystyle (1-x)^{-2} = \sum_{n=0}^\infty (n+1)x^n\), \(\displaystyle (1-x)^{-3} = \sum_{n=0}^\infty \frac{(n+2)(n+1)}{2}x^n\) \begin{align*} && \sum_{n=1}^{\infty} n2^{-n} &= \frac12\sum_{n=0}^{\infty}(n+1)2^{-n} \\ &&&= \frac12 (1-\tfrac12)^{-2} = 2 \\ \\ && \sum_{n=1}^{\infty} nx^n&= x(1-x)^{-2} \\ \Rightarrow && \sum_{n=1}^{\infty} n^2x^{n-1}&= (1-x)^{-2}+2x(1-x)^{-3} \\ \Rightarrow && \sum_{n=1}^{\infty} n^22^{-n} &= \frac12 \left ( (1-\tfrac12)^{-2}+2\cdot \tfrac12 \cdot (1-\tfrac12)^{-3} \right) \\ &&&= \frac12 \left ( 4 +8\right) = 6 \end{align*}
  2. By the generalised binomial theorem, \begin{align*} && (1-x)^{-\frac12} &= 1 + \sum_{n=1}^{\infty} \frac{(-\tfrac12)\cdot(-\tfrac32)\cdots(-\tfrac12-n+1)}{n!}(-x)^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{(-1)^n(\tfrac12)\cdot(\tfrac32)\cdots(\tfrac{2n-1}2)}{n!}(-x)^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^nn!}x^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{2^nn! \cdot 2^n n!}x^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\ &&&= \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\ \end{align*} \begin{align*} && \sum_{n=0}^\infty \frac{(2n)!} {(n!)^2 2^{2n}3^{n}} &= (1-\tfrac13)^{-\frac12} \\ &&&= \sqrt{\frac32} \\ \\ && (1-x)^{-\frac12} &= \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\ \Rightarrow && \tfrac12(1-x)^{-\frac32} &= \sum_{n=0}^{\infty} \frac{n(2n)!}{2^{2n}(n!)^2}x^{n-1} \\ \Rightarrow && \sum_{n=1}^\infty \frac{n(2n)!} {(n!)^2 2^{2n}3^{n}} &= \frac16(1-\tfrac13)^{-3/2} \\ &&&= \frac16 \sqrt{\frac{27}{8}} = \frac14\sqrt{\frac{3}2} \end{align*}

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2004 Paper 1 Q7
D: 1500.0 B: 1500.0
generalised binomial theorem recurrence relation power series partial fractions generating function geometric series linear recurrence

  1. The function \(\f(x)\) is defined for \(\vert x \vert < \frac15\) by \[ \f(x) = \sum_{n=0}^\infty a_n x^n\;, \] where \(a_0=2\), \(a_1=7\) and \(a_n =7a_{n-1} - 10a_{n-2}\) for \(n\ge{2}\,\). Simplify \(\f(x) - 7x\f(x) + 10x^2\f(x)\,\), and hence show that \(\displaystyle\f(x) = {1\over 1-2x} + {1 \over 1-5x} \;\). Hence show that \(a_n=2^n + 5^n\,\).
  2. The function \(\g(x)\) is defined for \(\vert x \vert < \frac13\) by \[ \g(x) = \sum_{n=0}^\infty b_n x^n \;, \] where \(b_0=5\,\), \(b_1 =10 \,\), \(b_2=40\,\), \(b_3=100\) and \(b_n = pb_{n-1} + qb_{n-2}\) for \(n\ge{2}\,\). Obtain an expression for \(\g(x)\) as the sum of two algebraic fractions and determine \(b_n\) in terms of \(n\).

Solution:

  1. \begin{align*} && f(x) -7xf(x)+10x^2f(x) &= \sum_{n=0}^\infty a_n x^n - 7x \sum_{n=0}^{\infty} a_n x^n + 10x^2 \sum_{n=0}^{\infty} a_nx^n \\ &&&= \sum_{n=2}^\infty (a_n-7a_{n-1}+10a_{n-2})x^n + a_0+a_1x-7a_0x \\ &&&= 0 + 2-7x \\ \\ \Rightarrow && f(x) &= \frac{2-7x}{1-7x+10x^2} \\ &&&= \frac{2-7x}{(1-5x)(1-2x)} \\ &&&= \frac{1}{1-2x} + \frac{1}{1-5x} \\ &&&= \sum_{n=0}^{\infty} (2^n + 5^n)x^n \end{align*} Therefore \(a_n = 2^n +5^n\)
  2. \(\,\) \begin{align*} && 40 &= 10p + 5 q \\ && 100 &= 40p+10q \\ && 10 &= 4p + q \\ \Rightarrow && (p,q) &= (1,6) \\ \\ && g(x) -xg(x)-6x^2g(x) &= 5+5x \\ \Rightarrow && g(x) &= \frac{5+5x}{1-x-6x^2} \\ &&&= \frac{5+5x}{(1-3x)(1+2x)} \\ &&&= \frac{4}{1-3x} + \frac{1}{1+2x} \\ &&&= \sum_{n=0}^{\infty} (4 \cdot 3^n + (-2)^n)x^n \\ \Rightarrow && b_n &= 4 \cdot 3^n + (-2)^n \end{align*}

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2003 Paper 1 Q5
D: 1500.0 B: 1500.6
binomial expansion generalised binomial theorem term independent of x binomial coefficient power series fractional powers

  1. In the binomial expansion of \((2x+1/x^{2})^{6}\;\) for \(x\ne0\), show that the term which is independent of \(x\) is \(240\). Find the term which is independent of \(x\) in the binomial expansion of \((ax^3+b/x^{2})^{5n}\,\).
  2. Let \(\f(x) =(x^6+3x^5)^{1/2}\,\). By considering the expansion of \((1+3/x)^{1/2}\,\) show that the term which is independent of \(x\) in the expansion of \(\f(x)\) in powers of \(1/x\,\), for \( \vert x\vert>3\,\), is \(27/16\,\). Show that there is no term independent of \(x\,\) in the expansion of \(\f(x)\) in powers of \(x\,\), for \( \vert x\vert<3\,\).

Solution:

  1. The terms will all be of the form \(x^{6-3k}\), so we are looking for \(\binom{6}{2}2^{4} \cdot 1^2 = 6 \cdot 5 \cdot 8 = 240\) The terms will be \(x^{(5n-k)3 -2k} = x^{15n-5k}\), so we want \(k = 3n\), \(\binom{5n}{3n}a^{5n-3n}b^{5n-2n} = \binom{5n}{3n}a^{2n}b^{3n}\)
  2. Let \(f(x) = (x^6+3x^5)^{1/2}\) If \(|x| > 3\), then consider \begin{align*} && f(x) &= x^3(1+3/x)^{1/2} \\ &&&= x^3 \left (1 + \frac12 \frac{3}{x} +\frac{1}{2!} \frac12\cdot \frac{-1}{2} \left ( \frac{3}{x} \right)^2 + \frac{1}{3!} \frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \left (\frac{3}{x} \right)^3 + \cdots \right) \\ &&&= \cdots \frac{1}{6} \frac{3}{8} 27x^0 + \cdots \\ &&&= \cdots + \frac{27}{16} + \cdots \end{align*} If \(|x| < 3\) we can consider \(f(0) = 0\) and notice that there must be no term independent of \(x\)

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1993 Paper 3 Q4
D: 1700.0 B: 1500.0
sequences infinite series summation power series partial fractions geometric series logarithmic series telescoping series differentiation of series

Sum the following infinite series.

  1. \[ 1 + \frac13 \bigg({\frac12}\bigg)^2 +\frac15\bigg(\frac12\bigg)^4 + \cdots + \frac{1}{2n+1} \bigg(\frac12\bigg)^{2n} + \cdots \] .
  2. \[ 2 -x -x^3 +2x^4 - \cdots + 2x^{4k} - x^{4k+1} - x^{4k+3} +\cdots \] where \(|x| < 1\).
  3. \[ \sum _{r=2}^\infty {r\, 2^{r-2} \over 3^{r-1} } \].
  4. \[ \sum_{r=2}^\infty {2 \over r(r^2-1) } \].

Solution:

  1. \begin{align*} && \sum_{i=0}^{\infty} x^{2i+1}&= \frac{x}{1-x^2} \\ \Rightarrow &&&=\frac12 \left ( \frac{1}{1-x} - \frac{1}{1+x} \right) \\ \underbrace{\Rightarrow}_{\int} && \sum_{i=0}^{\infty} \frac{1}{2i+1} x^{2i+2} &= \frac12 \left ( -\ln (1-x) - \ln(1+x) \right) \\ \underbrace{\Rightarrow}_{x = 1/2} && \sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i+2} &= -\frac12 \ln \frac12 - \frac12 \ln \frac32 \\ &&&= -\frac12 \ln \frac34 \\ &&\frac14\sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i}&= \frac12 \ln \frac43 \\ \Rightarrow&& S &= 2 \ln \frac43 \end{align*}
  2. \begin{align*} \sum_{k=0}^{\infty} \left (2x^{4k} - x^{4k+1} - x^{4k+3} \right) &= \sum_{k=0}^{\infty} \left (2 - x^{1} - x^{3} \right) x^{4k} \\ &= \frac{2-x-x^3}{1-x^4} \\ &= \frac{(1-x)(2+x+x^2)}{(1-x)(1+x+x^2+x^3)} \\ &= \frac{2+x+x^2}{1+x+x^2+x^3} \end{align*}
  3. \begin{align*} && \frac{1}{(1-x)^2} &= \sum_{r=0}^{\infty} r x^{r-1} \\ \Rightarrow && 9 &= \sum_{r=1}^{\infty} r \left ( \frac23 \right)^{r-1} \\ \Rightarrow && \sum_{r=2}^{\infty} r \left ( \frac{2^{r-2}}{3^{r-1}} \right) &= \frac12 \left ( 9 - 1 \right) \\ &&&= 4 \end{align*}
  4. \begin{align*} && \frac{2}{r(r^2-1)} &= \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \\ \Rightarrow && \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \right) &= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} - \frac{1}{r} + \frac{1}{r+1} \right) \\ &&&= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} \right)-\sum_{r=2}^{\infty} \left ( \frac{1}{r} - \frac{1}{r+1} \right) \\ &&&= 1 - \frac12 \\ &&&= \frac12 \end{align*}

View
1989 Paper 2 Q9
D: 1600.0 B: 1515.3
first order differential equations integrating factor matrix matrix exponential hyperbolic functions power series system of ODEs

The matrix \(\mathbf{F}\) is defined by \[ \mathbf{F}=\mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}, \] where $\mathbf{A}=\begin{pmatrix}-3 & -1\\ 8 & 3 \end{pmatrix} \( and \) t \( is a variable scalar. Evaluate \)\mathbf{A}^{2}$, and show that \[ \mathbf{F}=\mathbf{I}\cosh t+\mathbf{A}\sinh t. \] Show also that \(\mathbf{F}^{-1}=\mathbf{I}\cosh t-\mathbf{A}\sinh t\), and that \(\dfrac{\mathrm{d}\mathbf{F}}{\mathrm{d}t}=\mathbf{FA}\). The vector $\mathbf{r}=\begin{pmatrix}x(t)\\ y(t) \end{pmatrix}$ satisfies the differential equation \[ \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}=\mathbf{0}, \] with \(x=\alpha\) and \(y=\beta\) at \(t=0.\) Solve this equation by means of a suitable matrix integrating factor, and hence show that \begin{alignat*}{1} x(t) & =\alpha\cosh t+(3\alpha+\beta)\sinh t\\ y(t) & =\beta\cosh t-(8\alpha+3\beta)\sinh t. \end{alignat*}

Solution: \begin{align*} \begin{pmatrix} -3 & -1 \\ 8 & 3 \end{pmatrix}^2 &= \begin{pmatrix} 9-8 & 3-3 \\ -24+24 & -8+9 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \mathbf{I} \end{align*} Therefore: \begin{align*} \mathbf{F} &= \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n} \\ &= \mathbf{I} + \sum_{n=1}^{\infty} \frac{1}{(2n)!}t^{2n} \mathbf{I} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}t^{2n+1} \mathbf{A} \\ &= \cosh t \mathbf{I} + \sinh t \mathbf{A} \end{align*} Notice that \begin{align*} \mathbf{F} (\mathbf{I}\cosh t-\mathbf{A}\sinh t) &= (\mathbf{I}\cosh t+\mathbf{A}\sinh t)(\mathbf{I}\cosh t-\mathbf{A}\sinh t) \\ &= \mathbf{I}^2 \cosh^2 t+\mathbf{A}(\sinh t \cosh t - \cosh t \sinh t) - \mathbf{A}^2\sinh^2 t \\ &= \mathbf{I} \cosh^2 t - \mathbf{I} \sinh^2 t \\ &= \mathbf{I} \end{align*} Therefore \(\mathbf{F}^{-1} = \mathbf{I}\cosh t-\mathbf{A}\sinh t\) \begin{align*} \frac{\d \mathbf{F}}{\d t} &= \frac{\d }{\d t} \left [ \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}\right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^n \\ &= \left ( \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^{n-1} \right) \mathbf{A} \\ &= \mathbf{F}\mathbf{A} \end{align*} \begin{align*} && \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}&=\mathbf{0} \\ \Rightarrow && \mathbf{F} \frac{\d \mathbf{r}}{\d t} + \mathbf{FAr} &= \mathbf{0} \\ && \frac{\d }{\d t} \left ( \mathbf{F} \mathbf{r}\right) &= 0 \\ \Rightarrow && \mathbf{Fr} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= \mathbf{F}^{-1}\mathbf{c} \\ &&&= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\mathbf{c} \\ t = 0: && \binom{\alpha}{\beta} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\binom{\alpha}{\beta} \\ &&&= \binom{\alpha \cosh t}{\beta \cosh t} - \binom{-3\alpha-\beta}{8\alpha + 3\beta}\sinh t \\ &&&= \binom{\alpha \cosh t + (3\alpha + \beta) \sinh t}{\beta \cosh t -(8\alpha + 3\beta)\sinh t} \end{align*} as required

View
1989 Paper 3 Q9
D: 1700.0 B: 1516.0
sequences sum to infinity power series arithmetico-geometric series logarithmic series binomial series series manipulation

Obtain the sum to infinity of each of the following series.

  1. \(1{\displaystyle +\frac{2}{2}+\frac{3}{2^{2}}+\frac{4}{2^{3}}+\cdots+\frac{r}{2^{r-1}}+\cdots;}\)
  2. \(1{\displaystyle +\frac{1}{2}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{2^{2}}+\cdots+\frac{1}{r}\times\frac{1}{2^{r-1}}+\cdots;}\)
  3. \({\displaystyle \dfrac{1\times3}{2!}\times\frac{1}{3}+\frac{1\times3\times5}{3!}\frac{1}{3^{2}}+\cdots+\frac{1\times3\times\cdots\times(2k-1)}{k!}\times\frac{1}{3^{k-1}}+\cdots.}\)
[Questions of convergence need not be considered.]

Solution:

  1. \begin{align*} && \frac1{1-x} &= \sum_{r=0}^{\infty} x^r \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{(1-x)^2} &= \sum_{r=0}^\infty rx^{r-1} \\ \underbrace{\Rightarrow}_{x = \frac12} && 4 &= \sum_{r=0}^{\infty} \frac{r}{2^{r-1}} \end{align*}
  2. \begin{align*} && \frac1{1-x} &= \sum_{r=1}^{\infty} x^{r-1} \\ \underbrace{\Rightarrow}_{\int} && -\ln (1-x) &= \sum_{r=1}^{\infty} \frac1{r} x^r \\ \underbrace{\Rightarrow}_{x = \frac12} && \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r}} \\ \Rightarrow && 2 \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r-1}} \\ \end{align*}
  3. \begin{align*} && (1-x)^{-1/2} &= 1 + \frac{(-\tfrac12)}{1!} (-x) +\frac{(-\tfrac12)(-\tfrac32)}{2!}(-x)^2 + \cdots \\ &&&= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{2^rr!} x^r \\ \underbrace{\Rightarrow}_{x = \frac23} && \sqrt{3} &= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^r} \\ &&&= 1 + \frac{1}{1!} \frac23 + \frac13 \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \Rightarrow && 3\sqrt{3}-5 &= \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\ \end{align*}

View
1988 Paper 1 Q5
D: 1500.0 B: 1487.0
generalised binomial theorem partial fractions geometric series power series generating function convergence interval algebraic identity

Given that \(b>a>0\), find, by using the binomial theorem, coefficients \(c_{m}\) (\(m=0,1,2,\ldots\)) such that \[ \frac{1}{\left(1-ax\right)\left(1-bx\right)}=c_{0}+c_{1}x+c_{2}x^{2}+\ldots+c_{m}x^{m}+\cdots \] for \(b\left|x\right|<1\). Show that \[ c_{m}^{2}=\frac{a^{2m+2}-2(ab)^{m+1}+b^{2m+2}}{(a-b)^{2}}\,. \] Hence, or otherwise, show that \[ c_{0}^{2}+c_{1}^{2}x+c_{2}^{2}x^{2}+\cdots+c_{m}^{2}x^{m}+\cdots=\frac{1+abx}{\left(1-abx\right)\left(1-a^{2}x\right)\left(1-b^{2}x\right)}\,, \] for \(x\) in a suitable interval which you should determine.

Solution: \begin{align*} \frac{1}{(1-ax)(1-bx)} &=\frac{1}{b-a} \l \frac{b}{1-bx}-\frac{a}{1-ax}\r \\ &= \frac{1}{b-a} \l \sum_{k=0}^{\infty} b(bx)^k-\sum_{k=0}^{\infty} a(ax)^k \r \\ &= \frac{1}{b-a} \sum_{k=0}^{\infty} \l b^{k+1} - a^{k+1} \r x^k \end{align*} Therefore \(\displaystyle c_m = \frac{b^{k+1}-a^{k+1}}{b-a}\). \begin{align*} c_m^2 &= \frac{(b^{m+1}-a^{m+1})^2}{(b-a)^2} \\ &= \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \end{align*} \begin{align*} \sum_{m=0}^{\infty} c_m x^m &= \sum_{m=0}^{\infty} \l \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m+2}}{(b-a)^2} \r x^m \\ &= \frac{1}{(b-a)^2} \l \sum_{m=0}^{\infty} a^{2m+2} x^m-2\sum_{m=0}^{\infty} (ab)^{m+1} x^m+\sum_{m=0}^{\infty} b^{2m+2} x^m \r \\ &= \frac{1}{(b-a)^2} \l a^2\sum_{m=0}^{\infty} a^{2m} x^m-2ab\sum_{m=0}^{\infty} (ab)^{m} x^m+b^2\sum_{m=0}^{\infty} b^{2m} x^m \r \\ &= \frac{1}{(b-a)^2} \l \frac{a^2}{1-a^2x^2} - \frac{2ab}{1-abx} + \frac{b^2}{1-b^2x^2}\r \\ &= \frac{1+ab}{(1-a^2x)(1-abx)(1-b^2x)} \end{align*} Where geometric series will converge if \(|a^2x| < 1, |b^2x| < 1, |abx| < 1\), ie \(|x| < \min (\frac{1}{a^2}, \frac{1}{b^2} )\)

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