Question

[Thi Thuý Ha]

Tính nhanh

1/3+1/3^2+1/3^3+...+1/3^2007+1/3^2008

Answer 1

[l҉o҉n҉g҉ d҉z҉]

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Rightarrow3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}\)

\(2A=3A-A\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{2007}}-\frac{1}{3^{2008}}\)

\(=1-\frac{1}{3^{2008}}\)

\(2A=1-\frac{1}{3^{2008}}\Rightarrow A=\frac{1-\frac{1}{3^{2008}}}{2}\)

Answer 2

[FL.Han_]

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(\Leftrightarrow2A=1-\frac{1}{3^{2008}}\)

\(\Leftrightarrow2A=\frac{3^{2008}-1}{3^{2008}}\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{3^{2008}}\div2\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{2.3^{2008}}\)