Sửa đề: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

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

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

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

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

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

Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>\(S<\frac{3}{16}\)

mà 3/16<3/15=1/5

nên S<1/5

Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{100}{3^{100}}\)

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

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

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

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

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

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

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

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

Ta có: \(4A=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>\(A<\frac{3}{16}\)

mà \(\frac{3}{16}<\frac{3.2}{16}=\frac15\)

nên \(A<\frac15\)

Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

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

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

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

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

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

Ta có: \(4S=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)

=>\(S<\frac{3}{16}\)

mà 3/16<3/15=1/5

nên S<1/5

Ta có:

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

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

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

Đặt \(E=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

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

\(3E-E=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(2E=3-\frac{1}{3^{99}}< 3\)

\(E< \frac{3}{2}\)

\(2D< \frac{3}{2}-\frac{1}{3^{100}}< \frac{3}{2}\)

\(D< \frac{3}{4}\)

Vậy...

Theo bài ta có:

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

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

\(=\frac{\left(1-\frac{100}{3^{100}}\right)+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)}{2}< \frac{1+\frac{1}{2}}{2}=\frac{3}{2}:2=\frac{3}{4}\)

Đpcm

chứng minh àk

> nhé bạn