Binomial Theorem 2 Question 10
13.
Prove that
$ \begin{aligned} & +(-1)^{k} \begin{array}{cc} n & n-k \\ k & 0 \end{array}=\begin{array}{l} n \\ k \end{array}(2003,4 \mathrm{M}) \end{aligned} $
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Solution:
- To show that
$ \begin{aligned} & 2^{k} \cdot{ }^{n} C_{0} \cdot{ }^{n} C_{k}-2^{k-1} \cdot{ }^{n} C_{1} \cdot{ }^{n-1} C_{k-1} \\ & \quad+2^{k-2} \cdot{ }^{n} C_{2} \cdot{ }^{n-2} C_{k-2}-\ldots+(-1)^{k}{ }^{n} C_{k}{ }^{n-k} C_{0}={ }^{n} C_{k} \end{aligned} $
Taking LHS
$2^{k} \cdot{ }^{n} C_{k}{ }^{n} C_{k}-2^{k-1} \cdot{ }^{n} C_{1} \cdot{ }^{n-1} C_{k-1}+\ldots+(-1)^{k} \cdot{ }^{n} C_{k} \cdot{ }^{n-k} C_{0}$
$=\sum_{r=0}^{k}(-1)^{r} \cdot 2^{k-r} \cdot{ }^{n} C_{r} \cdot{ }^{n-r} C_{k-r}$
$=\sum_{r=0}^{k}(-1)^{r} 2^{k-r} \cdot \frac{n !}{r !(n-r) !} \cdot \frac{(n-r) !}{(k-r) !(n-k) !}$
$=\sum_{r=0}^{k}(-1)^{r} \cdot 2^{k-r} \cdot \frac{n !}{(n-k) ! \cdot k !} \cdot \frac{k !}{r !(k-r) !}$
$=\sum_{r=0}^{k}(-1)^{r} \cdot 2^{k-r}{ }^{n} C_{k} \cdot{ }^{k} C_{r}=2^{k} \cdot{ }^{n} C_{k} [\sum_{r=0}^{k}(-1)^{r} \cdot \frac{1}{2^{r}} \cdot{ }^{k} C_{r}]$
$=2^{k} \cdot{ }^{n} C_{k} (1-\frac{1}{2})^{k} \stackrel{ }{n} C_{k}=\mathrm{RHS}$
