Question

In the expansion of $(1+x+x^2+x^3+.....$ to $\infty {)}^3$, coefficient of $x^3$ is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

Answer: (D)

$10$

The expression $1+x+x^2+...\infty$

forms GP. so applying sum of GP to infinite

terms we have

$1+x+x^2+...=\frac{1}{1-x}$ $[\because S_{\infty }=\frac{a}{1-r}]$

So,

$(1+x+x^2+...{)}^3=(1-x{)}^{-3}$

Applying Binomial Theorem for

negative index we have

$(1-x{)}^{-3}=1{+}^3{C}_{1}x{+}^4{C}_2{x}^2{+}^5{C}_3{x}^3+...$

$\Rightarrow$ coeff. of $x^3$ is ${}^5{C}_3=10$

$\Rightarrow$ Option D is correct