Question

A=\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\)

Answer 1

\(A=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{340}\)

\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

Answer 2

\(\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)

= \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{20}\right)\)

= \(\frac{3}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

= 1. \(\frac{9}{20}\)

= \(\frac{9}{20}\)