Question

Rút gọn \(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{8.3^{99}}\)

Answer 1

Đặt D=\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

=>3D=\(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

=>3D-D=(\(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\))-(\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\))

=>2D=\(1-\dfrac{1}{3^{99}}\)

=>D=\(\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)

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