Question

1/2+1/4+1/8+1/16+1/32+...+1/2048+1/4096

Answer 1

Đặt A=1/2+1/4+1/8+1/16+1/32+...+1/2048+1/4096

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{12}}\)

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^{11}}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)

\(A=1-\frac{1}{2^{12}}\)