Question

$\underset{x\to 2}{lim}\left(\sum _{n=1}^9\frac{x}{n(n+1)x^2+2(2n+1)x+4}\right)$ is equal to:

• A. $\frac{7}{36}$
• B. $\frac{1}{5}$
• C. $\frac{5}{24}$
• D. $\frac{9}{44}$

Answer 1

The correct option is D $\frac{9}{44}$

$\underset{x\to 2}{lim}\left(\sum _{n=1}^9\frac{x}{n(n+1)x^2+2(2n+1)x+4}\right)$
$\Rightarrow \sum _{n=1}^9\frac{2}{4n^2+12n+8}$
$\Rightarrow \frac{1}{2}\sum _{n=1}^9\frac{1}{n^2+3n+2}$
$\Rightarrow \frac{1}{2}\sum _{n=1}^9\frac{(n+2)-(n+1)}{(n+1)(n+2)}$
$\Rightarrow \frac{1}{2}\sum _{n=1}^9(\frac{1}{n+1}-\frac{1}{n+2})$
$\Rightarrow \frac{1}{2}(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{10}-\frac{1}{11})$
$\Rightarrow \frac{1}{2}\left(\frac{9}{22}\right)=\frac{9}{44}$