Question

Tính A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

Answer 1

\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\)
\(A=\left(a-\frac{1}{2}A\right).2=\left(\frac{1}{2}-\frac{1}{2^{11}}\right).2=\frac{1023}{1024}\)

Answer 2

\(A=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\)

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

\(2A-A=1-\frac{1}{2^{10}}\)

Answer 3

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

A = 2A-A = ( 1+\(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ .....+\(\frac{1}{2^9}\)) - ( \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+......+\(\frac{1}{2^{10}}\)) 

A = 1-\(\frac{1}{2^{10}}\)

ok xong r nhé