Question

[Jerry Gaming]

Tính bằng cách thuận tiện nhất :

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

Answer 1

[Đàm Thị Minh Hương]

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)\(\Leftrightarrow A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}\)

Answer 2

[kudo shinichi]

nhân A với 1/2

ta có: A= 1/2+1/2^2+1/2^3+...+1/2^10

1/2.A= 1/2^2+1/2^3+...+1/2^11

lấy A-1/2.A= 1/2-1/2^11

1/2.A=1/2-1/2^11

A= (...-...):1/2