Question

tính

A= \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + ... + \(\frac{1}{512}\) + \(\frac{1}{1024}\) + \(\frac{1}{2048}\)

Answer 1

Ta có :\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}+\frac{1}{2048}\)

nên \(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\)

Do đó :  \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}+\frac{1}{2048}\right)\)

\(A=1-\frac{1}{2048}=\frac{2047}{2048}\)

Nhớ k giùm mình nhớ

Answer 2

ket qua là 2047\2048