Question

The value of $\underset{x\to 1}{lim}\frac{x+x^2...+x^n-n}{x-1}$ is

• A. n
• B. $\frac{n+1}{2}$
• C. $\frac{n(n+1)}{2}$
• D. $\frac{n(n-1)}{2}$

Answer 1

Answer: (C)

$\frac{n(n+1)}{2}$

$\underset{x\to 1}{lim}\frac{x+x^2...+x^n-n}{x-1}\left(\frac{0}{0}form\right)$
(L'Hospital's Rule)

$=\underset{x\to 1}{lim}\frac{x+2x+...+n{x}^{n-1}}{1}=1+2+...+n$
$=\frac{n(n+1)}{2}$
Hence, option 'C' is correct.