Đề bài: Tính
\(A=\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\frac{1}{512}+\frac{1}{2048}\)
\(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}\)
\(2^2.A=2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\)
\(4A-A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}+\frac{1}{2^{11}}\right)\)
\(3A=2-\frac{1}{2^{11}}\)
\(\Rightarrow A=\frac{2-\frac{1}{2^{11}}}{3}\)
Vậy \(A=\frac{2-\frac{1}{2^{11}}}{3}\).
ta có
A= 1/2+ 1/8+1/32+1/128+1/512+1/2048
=> A= 1/2 +1/ 2^3 +1/2^5 +1/2^7+1/2^9+1/2^11
=> 2^2 A=2+1/2+1/2^3+1/2^5+1/2^7+1/2^9
=> 2^2A-A= (2+1/2+1/2^3+1/2^5+1/2^7+1/2^9)-(1/2+1/2^3+/2^5+1/2^7+1/2^9+1/2^11)
=> 3A= 2- 1/2^11
=>3A= 4095/2048
=> A= 1365/2048
