Đặt \(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^{100}} \)
\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^{99}}\\ 2A-A=\left(2-1\right)+\dfrac{3}{2^2}+\left(\dfrac{4}{2^3}-\dfrac{3}{2^3}\right)+...+\left(\dfrac{100}{2^{99}}-\dfrac{99}{2^{99}}\right)+\dfrac{100}{2^{100}}\\ A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{100}{2^{100}}\\ 2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{100}{2^{99}}\\ 2A-A=2+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}\\ A=\dfrac{2^{100}+98}{2^{100}}\)