Question

Cách tính:1/2+1/4+1/8+...+1/64+1/128 Giúp mình với.

Answer 1

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\right)\)

\(A=1-\dfrac{1}{128}=\dfrac{127}{128}\)