A=1/(2x2)+1/(3x3)+...+1/(100x100) 
Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) 
=> A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4

A=1/(2x2)+1/(3x3)+...+1/(100x100) Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) => A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4

ra ba\o nhyieu

Bài 1: CMR:1/3+1/7+1/13+1/21+1/31+1/43+1/57+1/73+1/91<1

                           Giải

Ta đặt M=1/3+1/7+1/13+1/21+1/31+1/43+1/57+1/73+1/91

Vậy M<1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90 

       M< 1/2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10

      M< (1-1/2) +(1/2-1/3) +(1/3-1/4) +(1/4-1/5) +(1/5-1/6) +(1/6-1/7) +(1/7-1/8) +(1/8-1/9) +(1/9-1/10)

     M< 1-1/10 < 9/10      (1)

     Vì 9/10 < 1    (2)

     Từ(1) và (2) ta có : 1/3+1/7+1/13+1/21+1/31+1/43+1/57+1/73+1/91<1

Bài 2:So sánh với 1:     1/4+1/9+1/16 + 1/25 +...+1/10000

                                                    Giải

Ta đặt M =1/4+1/9+1/16 + 1/25 +...+1/10000

Hay M = 1/2X2+ 1/3X3+1/4X4+1/5X5 +...+1/100X100

        M< 1/1x2+ 1/2x3+1/3x4+1/4x5+...+1/99x100

        M< (1-1/2) +(1/2-1/3) +(1/3-1/4) +(1/4-1/5)+...+(1/99-1/100)

        M< 1-1/100 < 99/100      (1)

        Vì 99/100 < 1    (2)

         Từ(1) và (2) ta có : 1/4+1/9+1/16 + 1/25 +...+1/10000 <1

\(M=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{4^2}< \dfrac{1}{3\cdot4};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

\(\Rightarrow M< \dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-1-\dfrac{1}{2}-...-\dfrac{1}{50}\\ =\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\left(50.số\right)=\dfrac{50}{50}=1\)

Vậy \(M< 1\)

Mình chỉ so sánh với 1 được thôi à :((