Question

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)

Answer 1

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)

\(=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2450}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)