Q77 / 200 · Manipal (MET) 2013 · Mathematics
The value of $\lim _{x \rightarrow 1} \frac{\left[\sum_{K=1}^{100}-x^K\right]-100}{x-1}$ is
- A$-5050$
- B$0$
- Correct answerC$5050$
- DNone of these
Solution
$\lim _{x \rightarrow 1} \frac{\left[\sum_{K=1}^{100} x^K\right]-100}{(x-1)}$
$\begin{aligned} & =\lim _{x \rightarrow 1} \frac{\left(x+x^2+x^3+\ldots+x^{100}\right)-100}{(x-1)} \\ & =\lim _{x \rightarrow 1} \frac{(x-1)+\left(x^2-1\right)+\left(x^3-1\right)+\ldots+\left(x^{100}-1\right)}{(x-1)}\end{aligned}$
$\begin{aligned}=\lim _{x \rightarrow 1}\left\{\left(\frac{x-1}{x-1}\right)+\left(\frac{x^2-1}{x-1}\right)+\left(\frac{x^3-1}{x-1}\right)\right. & \\ & \left.+\ldots+\left(\frac{x^{100}-1}{x-1}\right)\right\}\end{aligned}$
$\begin{aligned}=\lim _{x \rightarrow 1}\left(\frac{x-1}{x-1}\right)+\lim _{x \rightarrow 1}\left(\frac{x^2-1}{x-1}\right) & +\lim _{x \rightarrow 1}\left(\frac{x^3-1}{x-1}\right) \\ & +\ldots+\lim _{x \rightarrow 1}\left(\frac{x^{100}-1}{x-1}\right)\end{aligned}$
$\begin{aligned} & =1+2+3+\ldots+100 \\ & =\Sigma 100=\frac{100 \times(100+1)}{2}\end{aligned}$
$=50 \times 101=5050 \quad\left\{\because \Sigma n=\frac{n(n+1)}{2}\right\}$.
Hence, the correct answer is option: 'C'.