Question

[Mori Ran]

N=1/3+1/3^2+1/3^3+...+1/3^9

mk cần gấp lắm các bn ơi, giúp mk nha

Answer 1

[Phạm Tuấn Đạt]

Ta có :\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)(1)

\(\Rightarrow3N=1+\frac{1}{3}+...+\frac{1}{3^8}\)(2)

Lấy (2) - (1) ta có :

\(\Rightarrow2N=1-\frac{1}{3^9}\)

\(\Rightarrow N=\frac{1-\frac{1}{3^9}}{2}\)

Answer 2

[I don't need you]

ta có: \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)

\(\Rightarrow\frac{1}{3}N=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{10}}\)

\(\Rightarrow\frac{1}{3}N-N=\frac{1}{3^{10}}-\frac{1}{3}\)

\(\frac{2}{3}N=\frac{1}{3^{10}}-\frac{1}{3}\)

\(N=\frac{\frac{1}{3^{10}}-\frac{1}{3}}{\frac{2}{3}}\)

CHÚC BN HỌC TỐT!!!