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2024 Paper 2 Q2
D: 1500.0 B: 1500.0
generalised binomial theorem integration partial fractions series expansion term-by-term integration logarithm trigonometric identities

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

  1. Find the binomial series expansion of \((8 + x^3)^{-1}\), valid for \(|x| < 2\). Hence show that, for each integer \(m \geqslant 0\), \[ \int_0^1 \frac{x^m}{8 + x^3}\,\mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{2^{3(k+1)}} \cdot \frac{1}{3k + m + 1} \right). \]
  2. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) = \int_0^1 \frac{1}{x+2}\,\mathrm{d}x\,, \] and evaluate the integral.
  3. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) = \pi\sqrt{a} - \ln b\,, \] where \(a\) and \(b\) are integers which you should determine.

Solution:

  1. Note that \(\,\) \begin{align*} && (8+x^3)^{-1} &= \tfrac18(1 + \tfrac18x^3)^{-1} \\ &&&= \tfrac18( 1 - \left (\tfrac{x}{2} \right)^3 + \left (\tfrac{x}{2} \right)^6 -\left (\tfrac{x}{2} \right)^9 + \cdots ) \end{align*} So \begin{align*} && \int_0^1 \frac{x^m}{8+x^3} \d x &= \int_0^1 x^m \sum_{k=0}^{\infty} \frac{1}{2^3} \left ( - \frac{x}{2} \right)^{3k} \d x \\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{1}{2^{3(k+1)}} \int_0^1 x^{m+3k} \d x \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \frac{1}{3k+m+1} \\ \end{align*}
  2. Notice that \begin{align*} && S &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) \\ &&&= \int_0^1 \frac{x^2}{8+x^3} \d x - 2\int_0^1 \frac{x^1}{8+x^3}+4\int_0^1 \frac{x^0}{8+x^3} \d x \\ &&&= \int_0^1 \frac{x^2-2x+4}{x^3+8} \d x \\ &&&= \int_0^1 \frac{1}{x+2} \d x = \left[ \ln(x+2)\right]_0^1 \\ &&&= \ln 3 - \ln 2 = \ln \tfrac32 \end{align*}
  3. Firstly, note that \begin{align*} && \frac{2k+1}{(3k+1)(3k+2)} &= \frac13 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \end{align*} so \begin{align*} && S_2 &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} 24 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \\ &&&= 24 \left [ \int_0^1 \frac{x^0}{8+x^3} \d x + \int_0^1 \frac{x^1}{8+x^3} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{1+x}{(x+2)(x^2-2x+4)} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{x+8}{12(x^2-2x+4)}-\frac{1}{12(x+2)} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x+8}{x^2-2x+4}-\frac{1}{x+2} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x-1}{x^2-2x+4}+ \frac{9}{(x-1)^2+3}-\frac{1}{x+2} \d x \right] \\ &&&=2 \left [ \int_0^1 \frac12\ln(x^2-2x+4)+3\sqrt{3}\arctan \frac{x-1}{\sqrt{3}} -\ln(x+2) \right]_0^1 \\ &&&=2 \left ( \frac12 \ln3+3\sqrt3 \arctan 0 - \ln 3 \right) - 2\left (\frac12 \ln 4-3\sqrt3 \arctan \frac{1}{\sqrt3} - \ln 2 \right) \\ &&&=\sqrt3 \pi - \ln3 \end{align*}

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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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2016 Paper 2 Q5
D: 1600.0 B: 1484.0
generalised binomial theorem binomial expansion combinatorial identity summation Vandermonde identity negative binomial proof binomial coefficients

In this question, the definition of \(\displaystyle\binom pq\) is taken to be \[ \binom pq = \begin{cases} \dfrac{p!}{q!(p-q)!} & \text{ if } p\ge q\ge0 \,,\\[4mm] 0 & \text{ otherwise } . \end{cases} \]

  1. Write down the coefficient of \(x^n\) in the binomial expansion for \((1-x)^{-N}\), where \(N\) is a positive integer, and write down the expansion using the \(\Sigma\) summation notation. By considering $ (1-x)^{-1} (1-x)^{-N} \, ,$ where \(N\) is a positive integer, show that \[ \sum_{j=0}^n \binom { N+j -1}{j} = \binom{N+n}{n}\,. \]
  2. Show that, for any positive integers \(m\), \(n\) and \(r\) with \(r\le m+n\), \[ \binom{m+n} r = \sum _{j=0}^r \binom m j \binom n {r-j} \,. \]
  3. Show that, for any positive integers \(m\) and \(N\), \[ \sum_{j=0}^n(-1)^{j} \binom {N+m} {n-j} \binom {m+j-1}{j } = \displaystyle \binom N n . \]

Solution:

  1. \(\frac{(-N)(-N-1)\cdots(-N-n+1)}{n!} = \binom{N+n-1}{n}\), so \[ (1-x)^{-N} = \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\] \begin{align*} && (1-x)^{-N-1} &= (1-x)^{-1}(1-x)^{-N} \\ &&&= (1 + x + x^2 + \cdots)\left ( \sum_{n=0}^{\infty} \binom{N+n-1}{n} x^n\right)\\ [x^{n}]: && \binom{N+1+n-1}{n} &= \sum_{j=0}^n \underbrace{1}_{x^{n-j} \text{ from 1st bracket}}\cdot\underbrace{\binom{N+j-1}{j}}_{x^j\text{ from second bracket}} \\ \Rightarrow && \binom{N+n}{n} &= \sum_{j=0}^n \binom{N+j-1}{j} \end{align*}
  2. Consider \((1+x)^{m+n} = (1+x)^m(1+x)^n\) and consider the coefficient of \(x^r\) from each side. On the left hand side this is clearly \(\binom{m+n}{r}\) on the right hand side we can take \(x^j\) from \((1+x)^m\) and \(x^{n-j}\) from \((1+x)^n\) and \(j\) can take any value from \(0\) to \(r\), ie \[ \binom{m+n} r = \sum _{j=0}^r \binom m j \binom n {r-j} \]
  3. Consider \((1-x)^{-(N+m+1)} = (\)

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2012 Paper 2 Q1
D: 1600.0 B: 1500.0
generalised binomial theorem power series expansion coefficient extraction generating functions partial fractions combinatorics geometric series

Write down the general term in the expansion in powers of \(x\) of \((1-x^6)^{-2}\,\).

  1. Find the coefficient of \(x^{24}\) in the expansion in powers of \(x\) of \[ (1-x^6)^{-2} (1-x^3)^{-1}\,.\] Obtain also, and simplify, formulae for the coefficient of \(x^n\) in the different cases that arise.
  2. Show that the coefficient of \(x^{24}\) in the expansion in powers of \(x\) of \[ (1-x^6)^{-2} (1-x^3)^{-1} (1-x)^{-1}\,\] is \(55\), and find the coefficients of \(x^{25}\) and \(x^{66}\).

Solution: \(\displaystyle (1-x^6)^{-2} = \sum_{n=0}^{\infty} (n+1)x^{6n}\)

  1. \(\,\) \begin{align*} && f(x) &= (1-x^6)^{-2}(1-x^3)^{-1} \\ &&&= \left ( \sum_{n=0}^{\infty} (n+1)x^{6n} \right) \left ( \sum_{n=0}^{\infty} x^{3n} \right) \\ [x^{24}]: && c_{24} &= 1 + 2+ 3+4+5 = 15 \end{align*} Clearly \(n\) must be a multiple of \(3\). If \(n = 6k\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) If \(n = 6k+3\) then we have \(1 + 2 + \cdots + (k+1) = \frac{(k+1)(k+2)}{2}\) the same way, we just must always get one extra \(x^3\) term from the second expansion.
  2. We can obtain \(x^{24}\) from the product of \((1-x^6)^{-2}(1-x^3)^{-1}\) and \((1-x)^{-1}\) in the following ways: \begin{array}{c|c|c} (1-x^6)^{-2}(1-x^3)^{-1} & (1-x)^{-1} & \text{product} \\ \hline 15x^{24} & x^0 & 15x^{24} \\ 10x^{21} & x^3 & 10x^{24} \\ 10x^{18} & x^6 & 10x^{24} \\ 6x^{15} & x^9 & 6x^{24} \\ 6x^{12} & x^{12} & 6x^{24} \\ 3x^{9} & x^{15} & 3x^{24} \\ 3x^{6} & x^{18} & 3x^{24} \\ x^{3} & x^{21} & x^{24} \\ x^{0} & x^{24}& x^{24} \end{array} So the total is \(55\). Similarly for \(25\) we can only obtain this in the same ways but also taking an extra power of \(x\) from the geometric series, ie \(55\) For \(66\) we obtain by similar reasoning that it is: \(\frac{13\cdot12}{2} + 2 \left (1 + 3 + \cdots + \frac{13 \cdot 12}{2} \right) = \frac{13\cdot12}{2} + 2 \binom{14}{3} = \frac{13 \cdot 12}2 ( 1 + \frac{30}{3}) = 11 \cdot 6 \cdot 13 = 858\)

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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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2007 Paper 2 Q1
D: 1600.0 B: 1516.0
generalised binomial theorem binomial expansion fractional exponent square root approximation cube root approximation numerical approximation series expansion

In this question, you are not required to justify the accuracy of the approximations.

  1. Write down the binomial expansion of \(\displaystyle \left( 1+\frac k {100} \right)^{\!\frac12}\)in ascending powers of \(k\), up to and including the \(k^3\) term.
    1. Use the value \(k=8\) to find an approximation to five decimal places for \(\sqrt{3}\,\).
    2. By choosing a suitable integer value of \(k\), find an approximation to five decimal places for \(\sqrt6\,\).
  2. By considering the first two terms of the binomial expansion of \(\displaystyle \left( 1+\frac k {1000} \right)^{\!\frac13}\), show that \(\dfrac{3029}{2100}\) is an approximation to \(\sqrt[3]{3}\).

Solution:

  1. Using the generalise binomial theorem \begin{align*} \left( 1+\frac k {100} \right)^{\frac12} &= 1 + \frac12 \frac{k}{100} + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)}{2!} \left (\frac{k}{100} \right)^2 + \frac{\tfrac12 \cdot \left ( -\tfrac12\right)\cdot \left ( -\tfrac32\right)}{3!} \left (\frac{k}{100} \right)^3 + \cdots \\ &= 1 + \frac{1}{200}k - \frac{1}{80\,000}k^2 + \frac{1}{16\,000\,000}k^3 + \cdots \end{align*}
    1. If \(k = 8\), \begin{align*} && \left( 1+\frac 8 {100} \right)^{\frac12} &= 1 + \frac{1}{200}8 - \frac{1}{80\,000}8^2 + \frac{1}{16\,000\,000}8^3 + \cdots \\ \Rightarrow && \frac{6\sqrt{3}}{10} &\approx 1 + 0.04 - 0.0008 + 0.000032 \\ &&&= 1.039232\\ \Rightarrow && \sqrt{3} &\approx 1.73205 \, (5\, \text{d.p.}) \end{align*}
    2. If \(k = -4\), \begin{align*} && \left( 1-\frac 4 {100} \right)^{\frac12} &= 1 - \frac{1}{200}4 - \frac{1}{80\,000}4^2 - \frac{1}{16\,000\,000}4^3 + \cdots \\ \Rightarrow && \frac{4\sqrt{6}}{10} &\approx 1 -0.02-0.0002 -0.000004 \\ &&&= 0.979796\\ \Rightarrow && \sqrt{6} &\approx 2.44949\, (5\, \text{d.p.}) \end{align*}
  2. \(\,\) \begin{align*} && \left( 1+\frac k {1000} \right)^{\!\frac13} &= 1 + \frac13 \frac{k}{1000} + \cdots \\ &&&= 1 + \frac{k}{3\,000} + \cdots \\ && 3 \times 7^3 &= 1029 \\ \Rightarrow && \left( 1+\frac {29} {1000} \right)^{\!\frac13} &\approx 1 + \frac{29}{3\,000} \\ \Rightarrow && \frac{7\sqrt[3]{3}}{10} &\approx \frac{3\,029}{3000} \\ \Rightarrow && \sqrt[3]{3} &= \frac{3\,029}{2\,100} \end{align*}

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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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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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2003 Paper 3 Q2
D: 1700.0 B: 1484.0
generalised binomial theorem binomial series combinatorial identity summation central binomial coefficient series convergence proof

Show that $\ds ^{2r} \! {\rm C}_r =\frac{1\times3\times\dots\times (2r-1)}{r!} \, \times 2^r \;, $ for \(r\ge1\,\).

  1. Give the first four terms of the binomial series for \(\l 1 - p \r^{-\frac12}\). By choosing a suitable value for \(p\) in this series, or otherwise, show that $$ \displaystyle \sum_{r=0}^\infty \frac{ {\vphantom {\A}}^{2r} \! {\rm C}_r }{ 8^r} = \sqrt 2 \; .$$
  2. Show that $$ \displaystyle \sum_{r=0}^\infty \frac{\l 2r + 1 \r \; {\vphantom{A}}^{2r} \! {\rm C} _r }{ 5^r} =\big( \sqrt 5\big)^3 \;. $$
[{\bf Note: } $ {\vphantom{A}}^n {\rm C}_r $ is an alternative notation for $\ds \ \binom n r \, \( for \)r\ge1\,\(, and \) {\vphantom{A}}^0 {\rm C}_0 =1 $ .]

Solution: \begin{align*} \binom{2r}{r} &= \frac{(2r)!}{r!r!} \\ &= \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots (2r-1)(2r)}{r! r!} \\ &= \frac{1 \cdot 3 \cdot 5 \cdots (2r-1) \cdot (2 \cdot 1) \cdot (2 \cdot 2) \cdots (2 \cdot r)}{r!}{r!} \\ &= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot 1 \cdot 2 \cdots r}{r!r!} \\ &= \frac{1\cdot 3 \cdots (2r-1) \cdot 2^r \cdot r!}{r!r!} \\ &= \frac{1\cdot 3 \cdots (2r-1)}{r!} \cdot 2^r \end{align*} which is what we wanted to show

  1. \begin{align*} (1 - p)^{-\frac12} &= 1 + \left ( -\frac12 \right )(-p) + \frac{1}{2!} \left (-\frac12 \right )\left (-\frac32 \right )(-p)^2 + \ldots \\ & \quad \quad \quad \cdots +\frac{1}{3!} \left (-\frac12 \right )\left (-\frac32 \right )\left (-\frac52 \right )(-p)^3 + O(p^4) \\ &= \boxed{1 + \frac{1}{2}p + \frac{3}{8}p^2 + \frac{5}{16}p^3} + O(p^4) \end{align*} More generally: \begin{align*} \binom{-\frac{1}{2}}{k} &=\frac{(-\frac{1}{2})\cdot(-\frac{1}{2} -1)\cdots(-\frac12 -k+1)}{k!} \\ &= \frac{(-1)(-3)(-5)\cdots(-(2k-1))}{k!2^k} \\ &= \frac{(-1)^k(1)(3)(5)\cdots((2k-1))}{k!2^k} \\ &= (-1)^k \frac{1}{4^k}\binom{2k}{k} \\ \end{align*} Therefore, \begin{align*} \sqrt{2} = \left (1-\frac12 \right)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-\frac12 \right )^r \tag{\(\frac12 < 1\) so series is valid} \\ &= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} \left (-\frac12 \right )^r \\ &= \sum_{r=0}^{\infty} \frac{1}{8^r}\binom{2r}{r} \end{align*}, which is what we wanted to show.
  2. \begin{align*} p(1-p^2)^{-\frac12} &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \left (-p^2 \right )^rp \\ &= \sum_{r=0}^{\infty} \frac{1}{4^r}\binom{2r}{r} p^{2r+1} \end{align*} Differentiating with respect to \(p\), \begin{align*} (1-p^2)^{-\frac12} +p^2(1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r} \\ (1-p^2)^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{4^r}(2r+1)\binom{2r}{r} p^{2r} \end{align*} Letting \(p = \frac{2}{\sqrt{5}}\), and \(|\frac2{\sqrt{5}}| < 1\) we have \begin{align*} \left (1-\frac45 \right )^{-\frac32} &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r} \\ (\sqrt{5})^3 &= \sum_{r=0}^{\infty} \frac{1}{5^r}(2r+1)\binom{2r}{r} \end{align*} (Alternative) \begin{align*} (\sqrt5)^3 &= \left ( \frac{1}{5} \right )^{-\frac32} \\ &= \left ( 1- \frac{4}{5} \right )^{-\frac32} \\ &= \sum_{r=0}^{\infty} \binom{-\frac32}{r} \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} \frac{-\frac32-(r-1)}{-\frac12} \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} \binom{-\frac12}{r} (2r+1) \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty} (-1)^r \frac{1}{4^r}\binom{2r}{r} (2r+1) \left (-\frac45 \right)^r \\ &= \sum_{r=0}^{\infty}(2r+1)\binom{2r}{r} \left (\frac15 \right)^r \\ (\sqrt{5})^3 &= \sum_{r=0}^{\infty}\frac{1}{5^r}(2r+1)\binom{2r}{r} \\ \end{align*}

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2001 Paper 2 Q1
D: 1600.0 B: 1500.0
generalised binomial theorem binomial expansion square root approximation polynomial approximation error estimation rational approximation small x expansion

Use the binomial expansion to obtain a polynomial of degree \(2\) which is a good approximation to \(\sqrt{1-x}\) when \(x\) is small.

  1. By taking \(x=1/100\), show that \(\sqrt{11}\approx79599/24000\), and estimate, correct to 1 significant figure, the error in this approximation. (You may assume that the error is given approximately by the first neglected term in the binomial expansion.)
  2. Find a rational number which approximates \(\sqrt{1111}\) with an error of about \(2 \times {10}^{-12}\).

Solution: \begin{align*} && \sqrt{1-x} &= (1-x)^{\frac12} \\ &&&= 1 -\frac12x+\frac{\frac12 \cdot \left (-\frac12 \right)}{2!}x^2 + \frac{\frac12 \cdot \left (-\frac12 \right) \cdot \left (-\frac32 \right)}{3!} x^3\cdots \\ &&&\approx 1-\frac12x - \frac18x^2 \end{align*}

  1. \(\,\) \begin{align*} && \frac{3\sqrt{11}}{10} &= \sqrt{1-1/100} \\ &&&\approx 1 - \frac{1}2 \frac{1}{100} - \frac{1}{8} \frac{1}{100^2} \\ &&&= \frac{80000-400-1}{80000} \\ &&&= \frac{79599}{80000}\\ \Rightarrow && \sqrt{11} &\approx \frac{79599}{24000} \\ \\ &&\text{error} &\approx \frac{1}{16} \frac{10}3 \frac{1}{100^3} \\ &&&= \frac{1}{48} 10^{-5} \\ &&&\approx 2 \times 10^{-7} \end{align*}
  2. Taking \(x = 1/10^4\) we have \begin{align*} && \frac{3 \sqrt{1111}}{100} &= \sqrt{1-1/10^4} \\ &&&\approx 1 - \frac12 \frac1{10^4} - \frac18 \frac{1}{10^8} \\ &&&= \frac{799959999}{800000000} \\ \Rightarrow && \sqrt{1111} & \approx \frac{266653333}{8000000} \\ \\ && \text{error} &\approx \frac{100}{3} \frac{1}{16} \frac{1}{10^{12}} \\ &&&= \frac{1}{48} \frac{1}{10^{10}} \\ &&&\approx 2 \times 10^{-12} \end{align*}

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1998 Paper 2 Q2
D: 1600.0 B: 1454.6
generalised binomial theorem binomial expansion approximation square root cube root fractional exponent numerical methods

Use the first four terms of the binomial expansion of \((1-1/50)^{1/2}\), writing \(1/50 = 2/100\) to simplify the calculation, to derive the approximation \(\sqrt 2 \approx 1.414214\). Calculate similarly an approximation to the cube root of 2 to six decimal places by considering \((1+N/125)^a\), where \(a\) and \(N\) are suitable numbers. [You need not justify the accuracy of your approximations.]

Solution: \begin{align*} && (1-1/50)^{1/2} &= 1 + \frac12 \cdot \left ( -\frac1{50} \right) + \frac1{2!} \frac12 \cdot \left ( -\frac12 \right)\cdot \left ( -\frac1{50} \right)^2 + \frac1{3!} \frac12 \cdot \left ( -\frac12 \right) \cdot \left ( -\frac32 \right)\cdot \left ( -\frac1{50} \right)^3 + \cdots \\ &&&=1-\frac{1}{100} - \frac12 \frac1{10000} -\frac12 \frac1{1000000} +\cdots \\ &&&= 0.9899495 + \cdots \\ \Rightarrow && \frac{7\sqrt{2}}{10} &\approx 0.9899495 \\ \Rightarrow && \sqrt{2} &\approx \frac{9.899495}{7} \\ &&&\approx 1.414214 \end{align*} \begin{align*} && (1 + 3/125)^{1/3} &= \frac{\sqrt[3]{125+3}}{5} \\ &&& = \frac{8\sqrt[3]{2}}{10} \\ && (1 + 3/125)^{1/3} &= 1 + \frac13 \left ( \frac{3}{125} \right) + \frac1{2!} \cdot \frac{1}{3} \cdot \left ( -\frac23\right) \left ( \frac{3}{125}\right)^2 +\cdots \\ &&&= 1+ \frac{8}{1000} - \frac{64}{1000000} \\ &&&= 1.007936 \\ \Rightarrow && \sqrt[3]{2} &= \frac{10.07936}{8} \\ &&&= 1.259920 \end{align*}

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1996 Paper 2 Q1
D: 1600.0 B: 1529.8
generalised binomial theorem binomial expansion negative exponent coefficient extraction polynomial powers series truncation

  1. Find the coefficient of \(x^{6}\) in \[(1-2x+3x^{2}-4x^{3}+5x^{4})^{3}.\] You should set out your working clearly.
  2. By considering the binomial expansions of \((1+x)^{-2}\) and \((1+x)^{-6}\), or otherwise, find the coefficient of \(x^{6}\) in \[(1-2x+3x^{2}-4x^{3}+5x^{4}-6x^{5}+7x^{6})^{3}.\]

Solution:

  1. We can obtain a \(6\) from \(4+2+0, 4+1+1, 3+3+0, 3+2+1, 2+2+2\). So \(x^6\) from \(4,2,0\) can happen in \(6\) ways and gets us a coefficient of \(1 \cdot 3 \cdot 5\). \(x^6\) from \(4,1,1\) can happen in \(3\) ways and gets us a coefficient of \(5 \cdot (-2) \cdot (-2)\). \(x^6\) from \(3,3,0\) can happen in \(3\) ways and gets us a coefficient of \((-4) \cdot (-4) \cdot 1\). \(x^6\) from \(3,2,1\) can happen in \(6\) ways and gets us a coefficient of \((-4) \cdot 3 \cdot (-2)\). \(x^6\) from \(2,2,2\) can happen in \(1\) ways and gets us a coefficient of \(3 \cdot 3 \cdot 3\). This leaves us with a total coefficient of: \(6 \cdot 15 + 3 \cdot 20 + 3 \cdot 16 + 6 \cdot 24 + 1 \cdot 27 = 369\)
  2. \begin{align*} (1+x)^{-2} &= 1 + (-2)x+\frac{(-2)\cdot(-3)}{2!} x^2 + \frac{(-2)(-3)(-4)}{3!}x^3 + \cdots \\ &= 1 -2x+3x^2-4x^3+5x^4+\cdots \\ \end{align*} The coefficient of \(x^6\) in the expansion of \((1+x)^{-6}\) will be \(\frac{(-6)(-7)(-8)(-9)(-10)(-11)}{6!} = \frac{11!}{6!5!} = 462\). The coefficient of \(x^6\) in the expansion of \((1 -2x+3x^2-4x^3+5x^4+\cdots)^3\) will be the same as the coefficient of \(x^6\) in the expansion of \((1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6)^3\), ie it will be \(462\)

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