Question

If $n$ is an even integer, then the value of ${}^n{C}_0{+}^n{C}_2{+}^n{C}_4+....{+}^n{C}_n$ equals to

• A. $2^n$
• B. $2^{n+1}$
• C. $2^{n-1}$
• D. $2^{2n}$

Answer 1

Answer: (C)

We know that,
${(1+x)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+....{+}^n{C}_n{x}^n$ ..(i)
and ${(1-x)}^n{=}^n{C}_0{-}^n{C}_{1}x{+}^n{C}_2{x}^2$
$+...+{(-1)}^n{C}_n{x}^n$ ..(ii)
On adding Eqs. (i) and (ii), we get
${(1+x)}^n+{(1-x)}^n=2{[}^n{C}_0{+}^n{C}_2{x}^2+....{+}^n{C}_n{x}^n]$
${}^n{C}_0{+}^n{C}_2+....{+}^n{C}_n=\frac{2^n}{2}$
$=2^{n-1}$