Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.....C_n{x}^n$, then $C_1+C_3+C_5........=?$

• A. $\frac{{(1+i)}^n-{(1-i)}^n+2^n}{2}$
• B. $\frac{{(1+i)}^n-{(1-i)}^n+2^n}{4}$
• C. $\frac{{(1+i)}^n+{(1-i)}^n+2^n}{2}$
• D. $\frac{{(1+i)}^n+{(1-i)}^n+2^n}{4}$

Answer 1

The correct option is D $\frac{{(1+i)}^n+{(1-i)}^n+2^n}{4}$

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)

Again,

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

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

$2^n=2C_1+2C_3+2C_5+........$

$2^{n-1}=C_1+C_2+C_3+C_5+........$

Hence, this is the answer.