Question

The value of $\underset{x\to 1}{\lim }\frac{\left[\sum _{K=1}^{100}-x^K\right]-100}{x-1}$ is

• A. -5050
• B. 0
• C. 5050
• D. None of these

Answer 1

Answer: (C)

$\underset{x\to 1}{\lim }\frac{\left[\sum _{k=1}^{100}{x}^k\right]-100}{(x-1)}$
$=\underset{x\to 1}{\lim }\frac{\left(x+x^2+x^3+\dots +x^{100}\right)-100}{(x-1)}$
$=\underset{x\to 1}{\lim }\frac{(x+1)+\left(x^2-1\right)+\left(x^3-1\right)+\dots +\left(x^{100}-1\right)}{(x-1)}$
$=\underset{x\to 1}{\lim }\left\{\left(\frac{x-1}{x-1}\right)+\left(\frac{x^2-1}{x-1}\right)+\left(\frac{x^3-1}{x-1}\right)+\dots +\left(\frac{x^{100}-1}{x-1}\right)\right\}$
$=\underset{x\to 1}{\lim }\left(\frac{x-1}{x-1}\right)+\underset{x\to 1}{\lim }\left(\frac{x^2-1}{x-1}\right)+\underset{x\to 1}{\lim }\left(\frac{x^3-1}{x-1}\right)$
$+\dots +\underset{x\to 1}{\lim }\left(\frac{x^{100}-1}{x-1}\right)$
$=1+2+3+\dots +100$
$=\Sigma 100=\frac{100\times (100+1)}{2}$
$=50\times 101=5050$
$\left\{\because \Sigma n=\frac{n(n+1)}{2}\right\}$