Question

rút gọn biểu thức: 

\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)

\(\frac{49}{58}.\frac{2^5}{4^2}-\frac{7^2}{-58}.3\)

Answer 1

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

\(\Rightarrow3D=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3D-D=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

\(\Leftrightarrow2D=1-\frac{1}{3^{100}}\)

\(\Leftrightarrow D=\frac{3^{100}-1}{2\cdot3^{100}}\)

Vậy \(D=\frac{3^{100}-1}{2\cdot3^{100}}\)

2) Ta có: \(\frac{49}{58}\cdot\frac{2^5}{4^2}-\frac{7^2}{-58}\cdot3\)

\(=\frac{49}{58}\cdot2-\frac{49}{58}\cdot3\)

\(=-1\cdot\frac{49}{58}\)

\(=-\frac{49}{58}\)