Mình nghĩ là 1

mik nghĩ vậy nhưng chưa bít trình bày í

Ta có: \(P=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(3P=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}\)

=>3P+P=\(1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2019}{3^{2018}}-\frac{2020}{3^{2019}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2019}{3^{2019}}-\frac{2020}{3^{2020}}\)

=>\(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

Đặt \(A=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>3A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}\)

=>3A+A=\(-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2018}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}\)

=>4A=-1\(-\frac{1}{3^{2019}}\)

=>\(4A=\frac{-3^{2019}-1}{3^{2019}}\)

=>\(A=\frac{-3^{2019}-1}{4\cdot3^{2019}}\)

Ta có: \(4P=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2019}}-\frac{2020}{3^{2020}}\)

\(=1+\frac{-3^{2019}-1}{4\cdot3^{2019}}-\frac{2020}{3^{2020}}=1+\frac{-3^{2020}-3-8080}{4\cdot3^{2020}}=1-\frac14-\frac{8083}{4\cdot3^{2020}}<\frac34\)

=>\(P<\frac{3}{16}\) (ĐPCM)

A=1/2x2/3x3/4x...x2018/2019x2019/2020=1/2020

A = 1/2 x 2/3 x 3/4 x ... x 2018/2019 x 2019/2020 = 1/2020

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