Question

[Đỗ Thị Thanh Lương]

Tính tổng sau:

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)

Answer 1

[Inami Sakura]

A=1/2+1/2^2+1/2^3+...+1/2^2016     (1)

2A=1+1/2+1/2^2+...+1/2^2015         (2)

Lấy (2)-(1) được:

A=1+1/2+1/2^2+...+1/2^2015-(1/2+1/2^2+1/2^3+...+1/2^2016)

A=1+1/2+1/2^2+...+1/2^2015-1/2-1/2^2-1/2^3-...-1/2^2016

A=1-1/2^2016

Vậy A=1-1/2^2016