Question

The value of the expression
$2^k(\frac{n}{0})(\frac{n}{k})-2^{k-1}(\frac{n}{1})(\frac{n-1}{k-1})+2^{k-2}(\frac{n}{2})(\frac{n-2}{k-2})..+(-1{)}^k(\frac{n}{k})(\frac{n-k}{0})$ is

• A. $(\frac{n}{k})$
• B. $(\frac{n+1}{k})$
• C. $(\frac{n+1}{k+1})$
• D. $(\frac{n-1}{k-1})$

Answer 1

The correct option is A $(\frac{n}{k})$

$2^k(\frac{n}{0})(\frac{n}{k})-2^{k-1}(\frac{n}{1})(\frac{n-1}{k-1})+2^{k-2}(\frac{n}{2})(\frac{n-2}{k-2})..+(-1{)}^k(\frac{n}{k})(\frac{n-k}{0})$
We know that

$(\frac{n}{0})(\frac{n}{k})=(\frac{k}{0})(\frac{n}{k})$

$(\frac{n}{1})(\frac{n-1}{k-1})=\frac{n!}{(n-1)!}\frac{(n-1)!}{(n-k)!(k-1)!}=\frac{k(n!)}{k!(n-k)!}=(\frac{k}{1})(\frac{n}{k})$

In the similar manner
$(\frac{n}{k})(\frac{n-k}{0})=(\frac{k}{k})(\frac{n}{k})$

$\Rightarrow 2^k(\frac{k}{0})(\frac{n}{k})-2^{k-1}(\frac{k}{1})(\frac{n}{k})+2^{k-2}(\frac{k}{2})(\frac{n}{k})..+(-1{)}^k(\frac{k}{k})(\frac{n}{k})$

$\Rightarrow (\frac{n}{k})[2^k(\frac{k}{0})-2^{k-1}(\frac{k}{1})+2^{k-2}(\frac{k}{2})...+(-1{)}^k(\frac{k}{k})]$

$\Rightarrow (\frac{n}{k}){(2-1)}^k$

$=(\frac{n}{k})$
Hence, option A.