Bài mình làm đơn giản thôi bạn nhé!

\(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+7}+...+\frac{1}{1+3+5+..2017}\)

Ta có: \(\frac{1}{1+3}< \frac{3}{4}\)

\(\frac{1}{1+3+5}< \frac{3}{4}\)

\(\frac{1}{1+3+5+7}< \frac{3}{4}\)

 .  .  .  . . . . .

\(\frac{1}{1+3+5+...+2017}< \frac{3}{4}\)

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\(A< \frac{3}{4}-\frac{1}{1+3+5+...+2017}\)

\(\Rightarrow A< \frac{3}{4}^{\left(đpcm\right)}\)

thằng tth quá ngu. làm vậy là sai bét.

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Nguyễn Phạm Nguyễn nói đúng oy. tth làm sai bét

\(A=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+2017}\)

\(\Rightarrow A=\frac{1}{\frac{\left(3+1\right).\left[\left(3-1\right):2+1\right]}{2}}+\frac{1}{\frac{\left(5+1\right).\left[\left(5-1\right):2+1\right]}{2}}+...+\frac{1}{\frac{\left(2017+1\right).\left[\left(2017-1\right):2+1\right]}{2}}\)

\(\Rightarrow A=\frac{1}{\frac{4.2}{2}}+\frac{1}{\frac{6.3}{2}}+...+\frac{1}{\frac{2018.1009}{2}}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1009^2}\)

à còn so sánh A với \(\frac{3}{4}\)nữa

A=1/2^2+1/3^2+...+1/1009^2

=>A<1/1.2+1/2.3+1/3.4+...+1/1008.1009

A<1-1/2+1/2-1/3+1/3-1/4+...+1/1008-1/1009

=>A<1-1/1009

=>A<3/4

\(1+3+5+7+....+\left(2n+1\right)=\left\{\left[\left(2n+1\right)-1\right]:2+1\right\}.\frac{2n+2}{2}=\left(n+1\right)^2\)

Áp dụng ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}\)

Ta có :\(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1009^2}< \frac{1}{1008.1009}\)

\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)

\(\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{1008}-\frac{1}{1009}=\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)

\(\Rightarrow A< \frac{3}{4}\left(đpcm\right)\)

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