\(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)
\(2A=1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\)
\(2A-A=\left(1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt:
\(B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(2B=2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(2B-B=\left(2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(B=2-\dfrac{1}{2^{99}}\)
Vậy \(A=2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}< 2\)
Bây giờ mình đang bận học bài 1 chút.Xíu nữa mình làm cho nhé
