Question

[Trần Đức Phát]

tính nhanh ;

A= 1/2+1/2^2+1/2^3+....+1/2^10

Answer 1

[Hotaru Takegawa]

Ta có:

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

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

\(\Rightarrow2A-A=A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)

Answer 2

[Nhớ đến phát điên mà vẫn thản nhiên...]

(^oy^o)