Question

Tính \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

Answer 1

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

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

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

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

=>\(A=\dfrac{3^{99}-1}{2\cdot3^{99}}\)