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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)

$\Rightarrow 1+2+3+\ldots+n=820$

$\Rightarrow \frac{n(n+1)}{2}=820$

$\Rightarrow n^{2}+n-1640=0$

$\Rightarrow n=40, n \in \mathrm{N}$

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