\(P=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>3P=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

=>3P+P=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{99}{3^{98}}-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>4P=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>4A=\(-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)

=>\(A=\frac{-3^{99}-1}{4\cdot3^{99}}\)

\(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}\)

=>4P<3/4

=>P<3/16

dốt thế 

Mình ngu lắm dân trần đăng ninh chuyên anh mà làm sao giỏi toán được

Đúng rồi đó ngu còn bày đặt

Lời giải:
Đặt \(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

\(3A=1-\frac{2}{3}+\frac{3}{3^2}-.....+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow 4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

$\Rightarrow 4A+12A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}<3$

$\Rightarrow 16A< 3$

$\Rightarrow A< \frac{3}{16}$

\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

=>\(\frac{1}{3}A=\frac{1}{3^2}-\frac{2}{3^3}+\frac{3}{3^4}-\frac{4}{3^5}+....+\frac{99}{3^{100}}-\frac{100}{3^{101}}\)

=>\(\frac{1}{3}A+A=\frac{4}{3}A=\frac{1}{3}-\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)+\left(\frac{4}{3^4}-\frac{3}{3^4}\right)+....+\left(\frac{99}{3^{99}}-\frac{98}{3^{99}}\right)+\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)

=>\(\frac{4}{3}A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+.....+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

Đặt \(S=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

=>\(\frac{1}{3}S=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+....+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

=>\(\frac{1}{3}S+S=\frac{4}{3}S=\frac{1}{3}-\frac{1}{3^{101}}\Rightarrow S=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{4}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{4}=\frac{1}{3}.\frac{3}{4}-\frac{1}{3^{101}}.\frac{3}{4}\)=>\(S=\frac{1}{4}-\frac{1}{3^{100}.4}\)

Mà \(\frac{4}{3}A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+....+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

=>\(\frac{4}{3}A=\frac{1}{4}-\frac{1}{3^{100}.4}=\frac{1}{4}-\frac{1}{3^{100}}.\frac{1}{4}=\frac{1}{4}.\left(1-\frac{1}{3^{100}}\right)\)

=>\(A=\frac{1}{4}\left(1-\frac{1}{3^{100}}\right):\frac{4}{3}=\frac{1}{4}\left(1-\frac{1}{3^{100}}\right).\frac{3}{4}=\frac{1}{4}.\frac{3}{4}.\left(1-\frac{1}{3^{100}}\right)=\frac{3}{16}.\left(1-\frac{1}{3^{100}}\right)\)

Vì \(1-\frac{1}{3^{100}}<1\Rightarrow A<\frac{3}{16}\)