Question

$\underset{x\to 1}{lim}\frac{x+x^2+...+x^n-n}{x-1}$ equals

• A. $n$
• B. $0$
• C. $\frac{n^2}{2}$
• D. $\frac{n(n+1)}{2}$

Answer 1

The correct option is D $\frac{n(n+1)}{2}$

Using L-hospital's rule,
$\underset{x\to 1}{lim}\frac{x+x^2+...+x^n-n}{x-1}$
$=\underset{x\to 1}{lim}\frac{1+2x+3x^2+...+n{x}^{n-1}}{1}$
$=1+2+3+.....................+n=\frac{n}{2}(2+(n-1\left).1\right)$
$=\frac{n(n+1)}{2}$
Hence, option 'D' is correct.