Question

$\underset{x\to 2}{lim}\frac{\sum _{r=1}^n{x}^r-\sum _{r=1}^n{2}^r}{x-2}=$

• A. $n$
• B. $(n-1)2^n$
• C. $(n-1)2^n+1$
• D. $(n-1)2^n-1$

Answer 1

The correct option is C $(n-1)2^n+1$

$\underset{x\to 2}{lim}\frac{\sum _{r=1}^n{x}^r-\sum _{r=1}^n{2}^r}{x-2}$

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

$=\underset{x\to 2}{lim}\frac{\frac{x(x^n-1)}{x-1}-2(2^n-1)}{x-2}$

It is of the form $\frac{0}{0}$. After applying L-Hospital's rule we get,

$=\underset{x\to 2}{lim}\frac{\frac{(x-1)\left(\right(n+1)x^n-1)-(x^{n+1}-x)}{{(x-1)}^2}}{1}$

$=(n-1)2^n+1$