Question

tính bằng cách thuận tiện      : \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}+\frac{1}{256}\)

Answer 1

Dễ lắm bạn à : 

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

 \(\Rightarrow2A=2\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{128}+\frac{1}{256}\right)\)

\(\Leftrightarrow2A=2+1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}+\frac{1}{128}\)

\(\Leftrightarrow2A-A=2+1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}+\frac{1}{128}-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{128}+\frac{1}{256}\right)\)

\(\Leftrightarrow A=2-\frac{1}{256}=\frac{511}{256}\)

Answer 2

đặt A= 1+1/2+1/4+1/8+...+1/128+1/256

2A=2+1+1/2+1/4+...+1/64+1/128

2A-A=A=2-1/256=511/256

Answer 3

ban giai ro ra duoc ko

Answer 4

giải thích cho mình đi