Question

${lim}_{x\leftarrow 1}\frac{x+x^2+x^3+....+x^n-n}{x-1}=$

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

Answer 1

Answer: (A)

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

$\underset{x\to 1}{lim}\frac{x+x^2+x^3+....+x^n-x}{x-1}\left(\frac{0}{0}\right)$ form

By L Hospital rule

$=\underset{x\to 1}{lim}\frac{1+2x+3x^2+...+n{x}^{n-1}}{1}$

$=1+2+3+...+n$

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