Question

Tính nhanh :

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

HELP!

Answer 1

Ta có :

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

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

Lấy (1) trừ (2) ta được :

\(2A=1-\frac{1}{3^8}=1-\frac{1}{6561}=\frac{6560}{6561}\)

Do đó : \(A=\frac{6560}{6561}:2=\frac{3280}{6561}\)

Vậy \(A=\frac{3280}{6561}\)

Answer 2

$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}$

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

$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}$