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)

$5050$

The value of $\underset{x\to 1}{lim}\frac{\left[\sum _{K=1}^{100}{x}^K\right]-100}{(x-1)}$ is
$\underset{x\to 1}{lim}\frac{(x+x^2+x^3+\cdots +x^{100})-100}{(x-1)}$
$=\underset{x\to 1}{lim}\frac{(x-1)+(x^2-1)+(x^3-1)+\cdots +(x^{100}-1)}{(x-1)}$
$=\underset{x\to 1}{lim}\{\left(\frac{x-1}{x-1}\right)+\left(\frac{x^2-1}{x-1}\right)+\left(\frac{x^3-1}{x-1}\right)+\cdots +\left(\frac{x^{100}-1}{x-1}\right)\}$
$=\underset{x\to 1}{lim}\left(\frac{x-1}{x-1}\right)+\underset{x\to 1}{lim}\left(\frac{x^2-1}{x-1}\right)+\underset{x\to 1}{lim}\left(\frac{x^3-1}{x-1}\right)+\cdots +\underset{x\to 1}{lim}\left(\frac{x^{100}-1}{x-1}\right)$
$=1+2+3+\cdots +100$
$=\sum 100=\frac{100\times (100+1)}{2}$
$=50\times 101=5050$ $\{\because \sum n=\frac{n(n+1)}{2}\}$