Question

Coefficient of $x^n$ in the expansion of $(1-2x+3x^2-4x^3+\dots \dots \infty {)}^2$ is

• A. $\frac{n(n+1)(n+2)}{3!}$
• B. $(-1)\frac{n(n+1)(n+2)}{3!}$
• C. $\frac{n(n+1)(n+2)(n+3)}{3!}$
• D. $\frac{(n+1)(n+2)(n+3)}{3!}$

Answer 1

The correct option is D $\frac{(n+1)(n+2)(n+3)}{3!}$

Let $S$ be $1-2x+3x^2-4x^3+......$

Consider $\int Sdx=\int (1-2x+3x^2-4x^3+.....)dx=(x-x^2+x^3-x^4+.....)$

$=\frac{x}{1-(-x)}$

$=\frac{x}{1+x}$ ($\because a+ar+a{r}^2+.....=\frac{a}{1-r}$)

$\Rightarrow S=\frac{d(\int Sdx)}{dx}=\frac{d\left(\frac{x}{1+x}\right)}{dx}=\frac{(1+x)\frac{d\left(x\right)}{dx}-\left(x\right)\frac{d(1+x)}{dx}}{(1+x{)}^2}=\frac{1+x-x}{(1+x{)}^2}=(1+x{)}^{-2}$

Thus, the given expansion becomes $(1+x{)}^{-2.2}=(1+x{)}^{-4}$

Coefficient of $x^n$ in the expansion of $(1+x{)}^{-4}$ is ${}^{(n+4-1)}{C}_{(4-1)}=\frac{(n+3)(n+2)(n+1)}{3!}$