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Q. The value of $\displaystyle\lim _{x \rightarrow 1} \frac{\left[\displaystyle\sum_{K=1}^{100}-x^{K}\right]-100}{x-1}$ is

ManipalManipal 2013

Solution:

$\displaystyle\lim _{x \rightarrow 1} \frac{\left[\displaystyle\sum_{k=1}^{100} x^{k}\right]-100}{(x-1)}$
$=\displaystyle\lim _{x \rightarrow 1} \frac{\left(x+x^{2}+x^{3}+\ldots+x^{100}\right)-100}{(x-1)}$
$=\displaystyle\lim _{x \rightarrow 1} \frac{(x+1)+\left(x^{2}-1\right)+\left(x^{3}-1\right)+\ldots+\left(x^{100}-1\right)}{(x-1)}$
$=\displaystyle\lim_{x \rightarrow 1} \left\{\left(\frac{x-1}{x-1}\right)+\left(\frac{x^{2}-1}{x-1}\right)+\left(\frac{x^{3}-1}{x-1}\right)+\ldots+\left(\frac{x^{100}-1}{x-1}\right)\right\}$
$=\displaystyle\lim _{x \rightarrow 1}\left(\frac{x-1}{x-1}\right)+\displaystyle\lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)+\displaystyle\lim _{x \rightarrow 1}\left(\frac{x^{3}-1}{x-1}\right)$
$+\ldots+\displaystyle\lim _{x \rightarrow 1}\left(\frac{x^{100}-1}{x-1}\right)$
$=1+2+3+\ldots+100$
$=\Sigma 100=\frac{100 \times(100+1)}{2}$
$=50 \times 101=5050$
$\left\{\because \Sigma n=\frac{n(n+1)}{2}\right\}$