Question

If $\underset{x\to 1}{lim}\frac{x+x^2+x^3+...+x^n-n}{x-1}=820,(n\in N)$ then the value of $n$ is equal to

Answer 1

$\underset{x\to 1}{lim}\frac{(x-1)}{x-1}+\frac{(x^2-1)}{x-1}+.....+\frac{(x^n-1)}{x-1}=820$
$\Rightarrow 1+2+3+......+n=820$
$\Rightarrow \sum n=820$
$\Rightarrow \frac{n(n+1)}{2}=820$
$\Rightarrow n=40$