Question

Tính tổng:

S=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+\dfrac{1}{3^5}+..+\dfrac{1}{3^9}\)

Answer 1

\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+................+\dfrac{1}{3^9}\)

\(\Rightarrow3S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+................+\dfrac{1}{3^8}\)

\(\Rightarrow3S-S=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+.............+\dfrac{1}{3^8}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+..........+\dfrac{1}{3^9}\right)\)

\(\Rightarrow2S=1-\dfrac{1}{3^9}\)

\(\Rightarrow S=\dfrac{1-\dfrac{1}{3^9}}{2}\)

Answer 2

Ta có : 3S = \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^8}\)

3S - S = \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^8}\) - \(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^9}\right)\)

2S = \(1-\dfrac{1}{3^9}=\dfrac{3^9-1}{3^9}\)

S = \(\dfrac{3^9-1}{2.3^9}\)

Answer 3

\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^9}\)(1)

nhân hai vế với 3, ta được:

\(3S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^8}\)(2)

trừ (1) cho (2), ta được:

\(3S-S=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^8}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^9}\right)\)\(2S=1-\dfrac{1}{3^9}\)

\(S=\dfrac{1-\dfrac{1}{3^9}}{2}\)

Answer 4

\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^9}\)

\(3S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^8}\)

\(2S=3S-S=1-\dfrac{1}{3^9}\)

\(\Rightarrow S=\dfrac{1-\dfrac{1}{3^9}}{2}\)

Vậy \(S=\dfrac{1-\dfrac{1}{3^9}}{2}\)

tik mik nha !!!