Question

\(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}\)

Answer 1

\(=\frac{1}{1x2}+\frac{1}{2x4}+\frac{1}{4x8} +\frac{1}{8x16}+\frac{1}{16x32}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)

\(=1-\frac{1}{32}\)

\(=\frac{31}{32}\)

nè ..!

Answer 2

\(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}\)

\(=\frac{256+64+16+4+1}{512}\)

\(=\frac{341}{512}\)