Question

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

Answer 1

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

A = [ 1 - \(\frac{1}{2}\)] + [ \(\frac{1}{2}\) - \(\frac{1}{4}\) ] + [\(\frac{1}{4}\) - \(\frac{1}{8}\)] + ..... + [\(\frac{1}{256}\) - \(\frac{1}{512}\)]

Xóa bỏ các số trùng lặp , ta có :

A = 1 - \(\frac{1}{512}\)

A = \(\frac{511}{512}\)

Answer 2

Tham khảo nha http://olm.vn/hoi-dap/question/98942.html

Answer 3

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{256}-\frac{1}{512}\)

\(=\frac{1}{1}-\frac{1}{512}\)

\(=\frac{511}{512}\)

Answer 4

\(A=\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(\Rightarrow2A=1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

\(\Rightarrow2A-A=A=1-\frac{1}{2^9}\)

\(=\frac{511}{512}\)