\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(-2A=2+1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{512}\)
\(-2A+A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)+\left(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(-A=2-\frac{1}{1024}\)
\(A=\frac{1}{2^{10}}-2\)
Đặt \(\begin{cases}A=-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{1024}\\B=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{1024}\end{cases}\)
=> B = - A
Ta có :
\(\left(2-1\right)B=\left(2-1\right)\left(1+\frac{1}{2}+....+\frac{1}{2^{10}}\right)\)
\(\Rightarrow B=\left(2+1+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+....+\frac{1}{2^{10}}\right)\)
\(\Rightarrow B=2-\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2^{10}}-2\)