Giải:
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{3}{2^4}+\dfrac{4}{2^5}+...+\dfrac{99}{2^{100}}+\dfrac{100}{2^{101}}\)
\(A-\dfrac{A}{2}=\dfrac{1}{2A}=\dfrac{1}{2}+\dfrac{3}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}-\dfrac{100}{2^{101}}\)
\(=\left[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right]-\dfrac{100}{2^{101}}\) ( Vì \(\dfrac{3}{2^3}=\dfrac{1}{2^2}+\dfrac{1}{2^3}\) )
\(=\dfrac{\left[1-\left(\dfrac{1}{2}\right)^{101}\right]}{\left(1-\dfrac{1}{2}\right)}-\dfrac{100}{2^{101}}\)
\(=\dfrac{\left(2^{101}-1\right)}{2^{100}}-\dfrac{100}{2^{101}}\)
\(\Rightarrow A=\dfrac{\left(2^{101}-1\right)}{2^{99}}-\dfrac{100}{2^{100}}\)
