Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.....C_n{x}^n$, then $C_0+C_2+C_4+C_6+........=?$

Answer 1

We have,

${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{x}^4+........+C_n{x}^n$ ……… (1)

On putting $x=1$ in equation (1), we get

$2^n=C_0+C_1+C_2+C_3+C_4+........+C_n$ ……… (2)

On putting $x=-1$ in equation (1), we get

$0=C_0-C_1+C_2-C_3+C_4-........$ ……… (3)

On adding equation (2) and (3), we get

$2^n=2C_0+2C_2+2C_4+........$

$2^{n-1}=C_0+C_2+C_4+C_6+........$

Hence, this is the answer.