Ta có : \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)

Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)

Nên \(A< B\)

\(\Rightarrow A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)

\(\Rightarrow A.B=\frac{1}{201}\)

Vì \(A< B\)

\(\Rightarrow A^2< A.B=\frac{1}{201}\)

\(\Rightarrow A^2< \frac{1}{201}\)

\(\RightarrowĐPCM\)

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)( đpcm )

A=[(3²-1)/3²].[(4²-1)/4²].[(5²-1)/5²] …[(50²-1)/50²] 
=(3-1)(3+1)(4-1)(4+1)(5-1)(5+1)…(50-1)(... /(3².4².5²…50²) 
= (3-1).(4-1).(5-1) … (50-1) .(3+1).(4+1).(5+1) … (50+1) (3².4².5²…50²) 
= 2.3.4 …49 . 4.5.6…51 /(3².4².5²…50²) 
=2.3. (4.5…49 . 4.5 … 49) . 50. 51 /(3².4².5²…50²) 
= 2.3.50.51(4².5²…49²)/(3².4².5²…50²) 
=2.3.50.51/(3².50²) 
=2.51/(3.50)=102/150=17/25 

2/Cho dãy số: 1(1/3); 1(1/8); 1(1/15); 1(1/24); 1(1/35); ... 
Có lẽ viết 1(1/3) là hỗn số tương đương với 4/3. 
a) Số hạng tổng quát : 1[1/[(n+1)²-1)] = (n+1)²/[(n+1)²-1]=(n+1)²/[n(n+1)] 
b) 
(đây là nghịch đảo của bài 1. Mẫu số phân tích tương tự tử số ở bài 1) 
Tích của 98 số hạng đầu là: 
P=[2²/(2²-1)].[3²/(3²-1)][4²/(4²-1)] …[99²/(99²-1)] 
= (2².3².4²…99²) /[(2²-1).(3²-1)… (99²-1)] 
= (2².3².4²…99²) /[(2-1).(3-1)… (99-1) . (2+1).(3+1)… (99+1)] 
= (2².3².4²…99²) /[1.2.3… 98 . 3.4… 98.99.100] 
= (2².3².4²…99²) /[1.2.(3… 98 . 3.4… 98).99.100] 
= (2².3².4²…99²) /[1.2.(3… 98 . 3.4… 98).99.100] 
= (2².3².4²…99²) /[1.2.(3… 98 . 3.4… 98).99.100] 
= (2².99²) /[1.2.99.100] 
=(2.99)/(1.100) 
=99/50 

3) 
C= (1/2).(3/4).(5/6).....(199/200). 
C= (1.3.5….199)/(2.4.6…200) 

C²= 1².3².5²….199²/(2².4².6²…200²) 
Ta có: k²>k²-1=(k-1)(k+1) nên 2²>1.3; 4²>3.5 … 200²>199.201. 
=> 
C² < 1².3².5²….199²/[(1.3).(3.5).(5.7)…(199.2...‡ 
=1².3².5²….199²/(1.3.3.5.5.7…199.201) 
=1².3².5²….199²/(1.3².5².7²…199².201) 
=1/201 

4) 
(cũng tương tự như bài 3) 
D= (1/2).(3/4).(5/6)…(99/100) 
D=(1.3.5..99)/(2.4.6…100) 
D²=(1².3².5²..99²)/(2².4².6²…100²) 

Làm nhỏ bớt mẫu số bởi: (k-1)(k+1)<k² 
D²=[(1².3².5²… 99²)]/(2².4².6²…100²) 
< 1².3².5²…99²/(1.3.3.5.5.7…99.01) 
=1².3².5²…99²/(1.3².5².7²…99².101) 
=1/101<1/100=1/10² 
=>D<1/10 

D²=(1².3².5²…99²)/(2².4².6²…100²) 
Giảm tử số bởi k²>(k-1)(k+1) 
D²=(1².3².5²..99²)/(2².4².6²…100²) 
>1².(2.4)(4.6)…(98.100) /(2².4².6²…100²) 
=2.4.4.6.6.8….96.98.98.100/(2².4².6²…10...‡ 
=2.4².6²…98².100/(2².4².6²…100²) 
=2.100/(2².100²) 
=1/200 > 1/225=1/15² 

=>D>1/15

ta có 1/2<2/3 ; 3/4<4/5;5/6<6/7;...;199/200<200/201

suy ra A^2=1/2^2*3/4^2*5/6^2*...*199/200^2<1/2*2/3*3/4*4/5*5/6*6/7*...*199/200/200/201

suy ra A^2<1/201(đpcm)

Ta có:

\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)

\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)

\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)

\(\Rightarrow A^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\right)\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\right)\)

\(\Rightarrow A^2< \frac{1}{201}\left(đpcm\right)\)

\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\)

\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}\frac{6}{7}...\frac{200}{201}\)

\(\Rightarrow A.A< A.\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)

\(\Rightarrow A^2< \frac{1}{201}\)(làm phần trc như Sakuraba Laura nhá)