Question

The co-efficent of $X^n$ in ${\left(1+X+2X^2+3X^3+....+n{X}^n\right)}^2$ is

• A. $\frac{n\left(n^2+11\right)}{6}$
• B. $\frac{n\left(n^2+10\right)}{6}$
• C. $\frac{n\left(n^2+11\right)}{4}$
• D. $\frac{n\left(n^2+10\right)}{4}$

Answer 1

The correct option is B $\frac{n\left(n^2+10\right)}{6}$

Let $s=(1+x+2x^2+3x^3...\dots +n{x}^n{)}^2$

$\Rightarrow s=1+x+2x^2+3x^3+...\dots +n{x}^n$

$\Rightarrow x.s=x^o+x^2+2x^3+3x^4+.....+n.x^{n+1}$

$\Rightarrow (1-x)s=1+x+x^2+x^3...\dots +x^n$

$\Rightarrow -n{x}^{n+1}-x=\frac{1-x^{n+1}}{1-x}-n{x}^{n+1}-x$

$\Rightarrow s=\frac{1}{(1-x{)}^2}-\frac{x}{1-x}=\frac{1-x+x^2}{(1-x{)}^2}$

Ignoring terms we have powers of x greater than $x^n$

coefficient of $x^n$ in

$(1+x+2x^2+3x^3+...\dots +n{x}^n{)}^2$

$\Rightarrow$ Coefficient of $x^n$ in $(1-x+x^2)(1-x^{-4})$

$\Rightarrow$ Such coefficient is clearly

$2n+\sum _{k=1}^{n-1}k(n-k)$

$\Rightarrow \frac{n(n^2+11)}{6}$

$\therefore$ So the answer is A$=\frac{n(n^2+11)}{6}$.