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