Question

[Nguyễn Tuyết Mai]

Tính nhanh (nếu có thể):

a, \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^8}\)

b, \(\frac{-1}{100.99}+\frac{-1}{99.98}+\frac{-1}{98.97}+..........+\frac{-1}{3.2}+\frac{-1}{2.1}\)

Answer 1

[Nguyễn Như Đạt]

b) \(GọiB=\frac{-1}{100.99}+\frac{-1}{99.98}+...+\frac{-1}{2.1}\)

\(2B=\frac{-2}{100.99}+\frac{-2}{99.98}+...+\frac{-2}{2.1}\)

\(2B=\frac{-1}{100}-\frac{-1}{99}+\frac{-1}{99}-\frac{-1}{98}+...+\frac{-1}{2}-\frac{-1}{1}\)

\(2B=\frac{-1}{100}-\frac{-1}{1}\)

\(2B=\frac{99}{100}\Rightarrow B=\frac{99}{100}:2=\frac{99}{200}\)

Answer 2

[Phan Nghĩa]

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(< =>3A=\frac{3}{3}+\frac{3}{3^2}+\frac{3}{3^3}+...+\frac{3}{3^8}\)

\(< =>3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(< =>3A-A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)

\(< =>A=\frac{3^8-1}{\frac{3^8}{2}}\)