Question

The sum of the series sigma $\mathrm{n}=8\mathrm{to}17$, $\frac{1}{(n+2)(n+3)}$ is equal to

• A. $\frac{1}{17}$
• B. $\frac{1}{18}$
• C. $\frac{1}{19}$
• D. $\frac{1}{20}$

Answer 1

The correct option is D

$\frac{1}{20}$

Explanation of the correct option:

Given statement can be written as,

$$\begin{gathered} \sum _{n=8}^{17}\frac{1}{\left(n+2\right)\left(n+3\right)} \\ =\sum _{n=8}^{17}\frac{1}{\left(n+2\right)}-\frac{1}{(n+3)} \\ =\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+..........+\frac{1}{19}-\frac{1}{20} \\ =\frac{1}{10}-\frac{1}{20} \\ =\frac{1}{20} \end{gathered}$$

Therefore the sum of the given series from $\mathrm{n}=8\mathrm{to}17$ is $\frac{1}{20}$.

Hence option D is correct .