Question

1,Tính nhanh

a A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{n-1}}+\dfrac{1}{3^n};n\in N\cdot\)

b, B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2007}}+\dfrac{1}{3^{2008}}\)

Answer 1

a: \(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{n+1}}\)

\(\Leftrightarrow-\dfrac{2}{3}A=\dfrac{1}{3^{n+1}}-\dfrac{1}{3}\)

hay \(A=\left(\dfrac{1}{3^{n+1}}-\dfrac{1}{3}\right):\dfrac{-2}{3}=\dfrac{1-3^n}{3^{n+1}}\cdot\dfrac{3}{-2}=\dfrac{3^n-1}{3^n\cdot2}\)

b: \(\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2009}}\)

\(\Leftrightarrow B\cdot\dfrac{-2}{3}=\dfrac{1}{3^{2009}}-\dfrac{1}{3}=\dfrac{1-3^{2008}}{3^{2009}}\)

\(\Leftrightarrow B=\dfrac{3^{2008}-1}{3^{2009}}:\dfrac{2}{3}=\dfrac{3^{2008}-1}{2\cdot3^{2008}}\)