Question

[what the fack]

Thuc hien phep tinh:

 B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005

Answer 1

[Cô nàng cự giải]

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)

\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)

\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)