Question

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\) =?

Answer 1

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{100}}\)

\(\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{101}}\)

\(\frac{1}{3}A-A=\)\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{101}}\)-\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{100}}\)

\(\frac{-2}{3}A=\frac{1}{3^{101}}-\frac{1}{3}\)

\(A=\frac{\frac{1}{3^{101}}-\frac{1}{3}}{\frac{-2}{3}}\)