Question

Cho A=\(\frac{4}{3}+\frac{10}{3^2}+\frac{28}{3^2}+...+\)\(\frac{3^{98}+1}{3^{98}}\).Chứng tỏ :98 < A < 99

 

Answer 1

A=\(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\)

=>  A>\(\frac{3}{3}+\frac{9}{9}+...+\frac{3^{98}}{3^{98}}\)  = 1+1+..+1 =98

A=\(\frac{3}{3}+\frac{9}{9}+...+\frac{3^{98}}{3^{98}}\) +\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)> 1+1+..+1 = 98 

Đặt  B = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

=> 3B  =  \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

=>2B =    1-\(\frac{1}{3^{98}}\)              <1    

=> B<1

=>A<99

=>98<A<99