Question

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

• A. $\frac{1}{17}$
• B. $\frac{1}{18}$
• C. $\frac{1}{19}$
• D. $\frac{1}{20}$
• E. $\frac{1}{21}$

Answer 1

Answer: (D)

$\frac{1}{20}$

${\sum }_{n=8}^{17}\frac{1}{(n+2)(n+3)}$

$={\sum }_{n=8}^{17}\frac{1}{(n+2)}-$$\frac{1}{(n+3)}$

$={\sum }_{n=8}^{17}\frac{1}{(n+2)}-$${\sum }_{n=8}^{17}\frac{1}{(n+3)}$

$=(\frac{1}{10}+\frac{1}{11}+.....+\frac{1}{19})-$$(\frac{1}{11}+\frac{1}{12}+.....+\frac{1}{20})$

$=\frac{1}{10}-\frac{1}{20}$

$=\frac{1}{20}$

Hence, D is the correct option.