Question

Cho M = \(\frac{4}{3}\) + \(\frac{10}{9}\) + \(\frac{28}{27}\) + ... + \(\frac{3^{98^{^{ }}}+1}{3^{98}}\) < 100

Answer 1

\(M=1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)

\(=1.98+\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

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

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

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

=>M=98+A<98+1<99<100

=>đpcm

Answer 2

chứng minh < 100 nha

 

Answer 3

Ê cái này ko có quy luật làm sao được