Đặt A la tên của biểu thức trên

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

\(=\frac{1}{2\left(3+1\right):2}+\frac{1}{3\left(5+1\right):2}+\frac{1}{4\left(7+1\right):2}+...+\frac{1}{1009\left(2017+1\right):2}\)

\(=\frac{2}{2.4}+\frac{2}{3.6}+\frac{2}{4.8}+....+\frac{2}{1009.2018}\)

\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{1009.1009}\)

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

Ta có: \(\frac{1}{2^2}=\frac{1}{4}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

...........

\(\frac{1}{1009^2}< \frac{1}{1008.1009}\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{1008.1009}\right)\)

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

\(A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{1009}\right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)

Vậy ...

Đặt tổng đã cho là A

\(\frac{1}{1+3}=\frac{1}{\left(3+1\right)x2:2}=\frac{1}{2x4:2}=\frac{1}{2x4}x2=\frac{2}{2x4}\)=\(\frac{1}{2x2}\)

\(\frac{1}{1+3+5}=\frac{1}{\left(1+5\right)x3:2}=\frac{1}{3x6}x2=\frac{2}{3x6}\)=\(\frac{1}{3x3}\)

\(\frac{1}{1+3+5+....+2017}=\frac{1}{\left(1+2017\right)x1009:2}=\frac{1}{1009x2018}x2=\frac{2}{1009x2018}\)=\(\frac{1}{1009x1009}\)

Các mẫu là bạn áp dụng tính tổng đó nha ( mk làm tắt)

A=\(\frac{1}{2x2}+\frac{1}{3x3}+...+\frac{1}{1009x1009}\)<\(\frac{1}{2x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{1008x1009}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1008}-\frac{1}{1009}\)=\(\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

vậy A<3/4( Mk có làm tắt nên chỗ nào ko hiểu thì nhắn tin nha

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A=1/(1+3)+1/(1+3+5)+1/(1+3+5+7)+...+1/(1+3+5+7+...+2017)

A=1/2^2+1/3^2+1/4^2+...+1/1009^2

2A=2/2^2+2/3^2+2/4^2+...+2/1009^2

Ta co :(x-1)(x+1)=(x-1)x+x-1=x^2-x+x-1=x^2-1<x^2

suy ra 2A<2/(1*3)+2/(3*5)+2/(5*7)+...+2/(1008*1010)

suy ra 2A <1-1/3+1/3-1/5+1/5-1/7+...+1/1008-1/1010

suy ra 2A<1-1/1010

suy ra 2A<2009/2010<1<3/2

suy ra 2A <3/2

suy ra A <3/4 (dpcm)

nho k cho minh voi nha

có cách nào dễ hiểu hơn không ạ?

A=1/(1+3)+1(1+3+5)+1/(1+3+5+7)+....+1/(1+3+5+7+...+2017)

A=1/4+1/9+1/16+....+1/1018081

A=1/2^2+1/3^2+1/4^2+...+1/1009^2

Ta có : 1/3^2=1/3x3<1/2x3

           1/4^2=1/4x4<1/3x4

           ......

           1/1009^2<1/1008x1009
   Suy ra 1/2^2+1/3^2+1/4^2+.....+1/1009^2<1/2^2+1/2x3+1/3x4+.....+1/1008x1009

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

              => A<1/2^2+1/2+1/3-1/3+......+1/1008-1/1008-1/1009

              => A<1/2^2+( 1/2-1/1009)

              => A< 3023/4036

Mà +) 3023<3/4
      +) A<3023/4026

Suy ra A<3/4

              => A<1008/1009

Ta có 1008/1009+

id nhu 1 tro dua

\(A=\left(-\frac{5}{11}\right).\frac{7}{15}+\frac{11}{-5}.\frac{30}{33}\)

\(A=-\frac{7}{33}+-2\)

\(A=-\frac{73}{33}\)

[ A] = -2

làm đc hết rùi phần b thui

\(D=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\left(\frac{1}{4}+\frac{3}{8}\right)\right]\)

\(D=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\frac{5}{8}\right]\)

\(D=\frac{1}{3}-\left(-\frac{15}{8}\right)\)

\(D=\frac{53}{24}\)

\(\left[D\right]=2\)

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

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

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

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

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