Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(B=2-\frac{1}{1024}=\frac{2047}{1024}\)
=> \(A=-\frac{2047}{1024}\)
\(-1-\frac{1}{2}-\frac{1}{4}-.....-\frac{1}{1024}\)
\(=-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\right)\)
Giả sử A\(=-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\right)\)
=> - 2A=\(=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\)
\(\Rightarrow-2A+A\)\(=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\) + \(\left[-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\right)\right]\)
\(\Rightarrow-A=1-\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2^{10}-1}\)