\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\Rightarrow2A=2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=\left(2+\frac{2}{3^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)
\(\Rightarrow A=\left(2-1\right)+\frac{2}{3^2}+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(\Rightarrow A=1+\frac{2}{3^2}+\left(\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(\Rightarrow A=1+\frac{2}{3^2}+\frac{1}{2}-\frac{1}{2^{98}}+\frac{100}{2^{100}}\)
Vậy...
