\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^5}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^4}\)
\(2A-A=A=1-\frac{1}{2^5}=1-\frac{1}{32}=\frac{31}{32}\)
đặt tổng là A
=>\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^5}\)
=>\(2A=1+\frac{1}{2}+...+\frac{1}{2^4}\)
=>\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^4}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^5}\right)=1-\frac{1}{2^5}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32
= 16/32 + 8/32 + 4/32 + 2/ 32
=\(\frac{16+8+4+2}{32}\)
= \(\frac{30}{32}\)
= \(\frac{15}{16}\)
