Question

Rút gọn \(C=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+...+\dfrac{1}{3^{99}}+\dfrac{1}{8.3^{99}}\)

Answer 1

Đặt :

\(A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+.......................+\dfrac{1}{3^{99}}+\dfrac{1}{3^{99}}\)

\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+...................+\dfrac{1}{3^{98}}\)

\(\Rightarrow3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+..............+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^3}+..............+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\right)\)\(\Rightarrow2A=1-\dfrac{1}{3^{99}}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)

\(\Rightarrow C=A+\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}+\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{3}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{8.3^{98}}=\dfrac{4.3^{98}-1}{8.3^{98}}\)