Question

$\underset{x\to 1}{lim}\frac{\sum _{r=1}^n{x}^r-n}{x-1}$ is equal to

• A. $\frac{n}{2}$
• B. $\frac{n(n+1)}{2}$
• C. $1$
• D. none of these

Answer 1

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

${lim}_{x\to 1}\frac{{\sum }_{r=1}^n{x}^r-n}{x-1}={lim}_{x\to 1}\frac{x^1+x^2+x^3+...+x^n-n}{x-1}={lim}_{x\to 1}\frac{x\frac{1-x^n}{1-x}-n}{x-1}={lim}_{x\to 1}\frac{x-x^{n+1}-n(1-x)}{-{(1-x)}^2}$
$={lim}_{x\to 1}\frac{1-(n+1)x^n+n}{2(1-x)}={lim}_{x\to 1}\frac{-(n+1)n{x}^{n-1}}{-2}$ (Applying L'Hospital rule)
$=\frac{n(n+1)}{2}$
Hence, option 'B' is correct.