Question

S=1/4+1/8+1/16+1/32+1/64+1/128

Answer 1

S=1/4+1/8+1/16+1/32+1/64+1/128

\(S=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)

\(2S=2\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

\(2S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)

\(2S-S=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

\(S=\frac{1}{2}-\frac{1}{2^7}\)

Answer 2

S=(1/2-1/4)+(1/4-1/8)+(1/8-1/16)+(1/16-1/32)+(1/32-1/64)+(1/64-1/128)

S=1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128

S=1/2-(1/4-1/4)+(1/8-1/8)+(1/16-1/16)+(1/32-1/32)+(1/64-1/64)-1/128

S=1/2-1/128

S=63/128