Question

${}^n{C}_0$ + ${}^n{C}_2$ + ${}^n{C}_4$...................=

• A. 0
• B. 1
• C.
• D.

Answer 1

The correct option is D

$(1+x{)}^n$ = ${}^n{C}_0$ + ${}^n{C}_1$x + ${}^n{C}_2$$x^2$ + ${}^n{C}_3$$x^3$ + ${}^n{C}_4$$x^4$..................

Put x = 1

⇒ $2^n$ = ${}^n{C}_0$ + ${}^n{C}_1$ + ${}^n{C}_2$ + ${}^n{C}_3$ + ${}^n{C}_3$ + ${}^n{C}_4$.................. (1)

Put x = (-1)

⇒ 0 = ${}^n{C}_0$ - ${}^n{C}_1$ + ${}^n{C}_2$ - ${}^n{C}_3$ + ${}^n{C}_4$......................... (2)

(1) + (2) ⇒ $2^n$ = 2[${}^n{C}_0$ + ${}^n{C}_2$ + ${}^n{C}_4$................]

⇒ ${}^n{C}_0$ + ${}^n{C}_2$ + ${}^n{C}_4$................= $\frac{2^n}{2}$

= $2^{n-1}$