Question

The sum of the series $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}$ is

• A. $\frac{19}{20}$
• B. $\frac{20}{21}$
• C. $\frac{20}{19}$
• D. $\frac{21}{20}$

Answer 1

The correct option is B $\frac{20}{21}$

$S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}{S}_n=\sum _{n=1}^n\frac{1}{n(n+1)}=\sum _{n=1}^n\frac{1}{n}-\frac{1}{n+1}=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n}-\frac{1}{n+1})S_n=1-\frac{1}{n+1}=\frac{n}{n+1}\text{As}420=20\times 21\Rightarrow n=20S_{20}=\frac{20}{21}$