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

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

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

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

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

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