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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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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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