Question

1/2+1/4+1/8....1/128+1/256

 

Answer 1

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}+\frac{1}{128}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\right)\)

\(A=1-\frac{1}{256}=\frac{255}{256}\)

Answer 2

\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ .... + \(\frac{1}{128}\)+ \(\frac{1}{256}\)

\(2A\)= \(1+\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)

\(2A-A\)= ( \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)) - ( \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ ...+ \(\frac{1}{128}\)+ \(\frac{1}{256}\))

\(A\)= 1- \(\frac{1}{256}\)

\(A\)= \(\frac{255}{256}\)

tk nhé

Answer 3

A = 255/256

tk nha bn