Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{512}\)
\(\Rightarrow2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\right)\)
\(\Rightarrow A=2-\dfrac{1}{1024}\)
Từ đó: \(-1-\dfrac{1}{2}-\dfrac{1}{4}-...-\dfrac{1}{1024}\)
\(=-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\right)\)
\(=-\left(2-\dfrac{1}{1024}\right)\)
\(=\dfrac{1}{1024}-2\)
Vậy...