Question:
If $\lim _{x \rightarrow 1} \frac{x+x^{2}+x^{3}+\ldots+x^{n}-n}{x-1}=820,(n \in \mathbf{N})$ then the value of $n$ is equal to
Solution:
$\lim _{x \rightarrow 1} \frac{x+x^{2}+x^{3}+\ldots .+x^{n}-n}{x-1}=820\left(\frac{0}{0}\right.$ case $)$
$\lim _{x \rightarrow 1} \frac{1+2 x+3 x^{2}+\ldots . .+n x^{n-1}}{1}=820$
(Using L' Hospital rule)
$\Rightarrow 1+2+3+\ldots+n=820$
$\Rightarrow \frac{n(n+1)}{2}=820$
$\Rightarrow n^{2}+n-1640=0$
$\Rightarrow n=40, n \in \mathrm{N}$