Question

B = 1 + \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{20}}\)

Answer 1

\(B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{20}}\)

\(\Rightarrow\dfrac{1}{3}B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{21}}\)

\(\Rightarrow\dfrac{1}{3}B-B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{21}}-1-\dfrac{1}{3}-...-\dfrac{1}{3^{20}}\)

\(\Rightarrow\dfrac{-2}{3}B=\dfrac{1}{3^{21}}-1\)

\(\Rightarrow B=\left(\dfrac{1}{3^{21}}-1\right):\left(-\dfrac{2}{3}\right)\)