Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.....C_n{x}^n$, then $C_0+C_4+C_8........=$

• 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)

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

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

$2^{n-1}=C_0+C_2+C_4+C_6+........$ …….. (4)

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

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

${(1+i)}^n=C_0+C_{1}i-C_2-C_{3}i+C_4+........$ ……. (5)

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

${(1-i)}^n=C_0-C_{1}i+C_2{(-i)}^2+C_3{(-i)}^3+C_4{(-i)}^4+........$

${(1-i)}^n=C_0-C_{1}i-C_2+C_{3}i+C_4+........$ …….. (6)

On adding equation (5) and (6), we get

${(1+i)}^n+{(1-i)}^n=2(C_0-C_2+C_4-C_6+C_8-........)$

$\frac{{(1+i)}^n+{(1-i)}^n}{2}=C_0-C_2+C_4-C_6+C_8-.......$ …….. (7)

On adding equation (4) and (7), we get

$\frac{{(1+i)}^n+{(1-i)}^n}{2}+2^{n-1}=2C_0+2C_4+2C_8........$

$\frac{{(1+i)}^n+{(1-i)}^n+2^n}{2}=2C_0+2C_4+2C_8........$

$C_0+C_4+C_8........=\frac{{(1+i)}^n+{(1-i)}^n+2^n}{4}$

Hence, this is the answer.