Question

Cho B = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ... + 1/3^2004 + 1/3^2005
Chứng minh rằng B < 1/2
 

Answer 1

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

=>2B=1-1/3^2005

=>\(2B=\dfrac{3^{2005}-1}{3^{2005}}\)

=>\(B=\dfrac{3^{2005}-1}{3^{2005}\cdot2}< \dfrac{1}{2}\)

Answer 2

         B  =      \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +........+ \(\dfrac{1}{3^{2024}}\)+ \(\dfrac{1}{3^{2005}}\)

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

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

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

           B  = ( 1 - \(\dfrac{1}{3^{2005}}\)):2

            B  =  \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{2005}}\) < \(\dfrac{1}{2}\) (đpcm)