Binomial Theorem 2 Question 10
13. Prove that
$$ \begin{aligned} & +(-1)^{k} \begin{array}{cc} n & n-k \\ k & 0 \end{array}=\begin{array}{l} n \\ k \end{array}(2003,4 M) \end{aligned} $$
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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
$2^{k} \cdot{ }^{n} C _k{ }^{n} C _k-2^{k-1} \cdot{ }^{n} C _1 \cdot{ }^{n-1} C _{k-1}+\ldots+(-1)^{k} \cdot{ }^{n} C _k \cdot{ }^{n-k} C _0$
$=\sum _{r=0}^{k}(-1)^{r} \cdot 2^{k-r} \cdot{ }^{n} C _r \cdot{ }^{n-r} C _{k-r}$
$=\sum _{r=0}^{k}(-1)^{r} 2^{k-r} \cdot \frac{n !}{r !(n-r) !} \cdot \frac{(n-r) !}{(k-r) !(n-k) !}$
$=\sum _{r=0}^{k}(-1)^{r} \cdot 2^{k-r} \cdot \frac{n !}{(n-k) ! \cdot k !} \cdot \frac{k !}{r !(k-r) !}$
$=\sum _{r=0}^{k}(-1)^{r} \cdot 2^{k-r}{ }^{n} C _k \cdot{ }^{k} C _r=2^{k} \cdot{ }^{n} C _k \sum _{r=0}^{k}(-1)^{r} \cdot \frac{1}{2^{r}} \cdot{ }^{k} C _r$
$=2^{k} \cdot{ }^{n} C _k 1-\frac{1}{2}^{k} \stackrel{ }{n} C _k=RHS$
