1 + 2 x 2 + 3 x 3 + 4 x 4 + 5 x 5 + 6 x 6 + 7 x 7 + 8 x 8 + 9 x 9 + 10 x 10

= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100

= 385

1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10

= 1+4+9+16+25+36+49+64+81+100

=(81+9)+(64+16)+(49+1)+)36+4)+25+100

=90+80+50+40+25 +100

=385

Đặt T = 1 + 2 x 2 + ... + 10 x 2 = 385

Ta thấy:

1/2*2<1/1*2)vì 2*2>1*2).

1/3*3<1/2*3(vì 3*3>2*3).

...

1/8*8<1/7*8(vì 8*8>7*8).

=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.

=>B<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.

=>B<1-1/8.

=>B<7/8.

Mà 7/8<1.

=>B<1.

Vậy B<1(đpcm).

\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(\Rightarrow1-\frac{1}{8}< 1\)

=>B<1

\(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6.}+\frac{1}{7.7}+\frac{1}{8.8}\)\(=\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(B=1-\frac{1}{8}\)

\(\Rightarrow B< 1\left(ĐPCM\right)\)

1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81 
10 x 10 = 100
 

\(1x1=1\)

\(2x2=4\)

\(3x3=9\)

\(4x4=16\)

\(5x5=25\)

\(6x6=36\)

\(7x7=49\)

\(8x8=64\)

\(9x9=81\)

\(10x10=100\)

1 x 1 = 1

2 x 2 = 4

3 x 3 = 9

4 x 4 = 16

5 x 5 = 25

6 x 6 = 36

7 x 7 = 49

8 x 8 = 64

9 x 9 = 81

10 x 10 = 100

a)=1x2x3x4x5/2x3x4x5x6 =1/6

b)=1x2x3x4x5x6x7x8x9/10x9x8x7x6x5x4x3 x2=1/10

1,4,9,16

ai đúng mình tích

1x1=1

2x2=4

3x3=9

4x4=16

5x5=25

6x6=36

7x7=49

8x8=64

9x9=81

10x10=100

11x11=121

12x12=144

13x13=169

14x14=196

15x15=225

16x16=256

17x17=289

18x18=324

19x19=361

20x20=400

\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}\)

\(\dfrac{1}{2.2}< \dfrac{1}{1.2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3.3}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{4.4}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{2009.2009}< \dfrac{1}{2008.2009}=\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1\)

Ta có:

\(\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{2009\times2009}< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2008\times2009}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)

 Ta có : 1/[n x (n - 1)] = [(n - 1) - n] / [n x (n - 1)] = 1/n - 1/(n - 1) 
Áp dụng : 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) 
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/48 - 1/49 + 1/49 - 1/50 
= 1 - 1/50 < 1 
Vậy : 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) < 1 
Ta có : 1/(n x n) < 1/[(n - 1) x n] 
1/(2x2) < 1/(1x2) 
1/(3x3) < 1/(2x3) 
1/(4x4) < 1/(3x4) 
............. 
1/(49x49) < 1/(49x49) 
1/(50x50) < 1/(49x50) 
=> 1/(2x2) + 1/(3x3) + 1/(4x4) + ... 1/(49x49) + 1/(50x50) < 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) < 1 
Vậy 1/(2x2) + 1/(3x3) + 1/(4x4) + ... 1/(49x49) + 1/(50x50) < 1

Đặt B=1/1*2+1/2*3+...+1/99*100 

Ta thấy:

A=1/2*2+1/3*3+...+1/100*100<B=1/1*2+1/2*3+...+1/99*100   (1)

Ta lại có: 

B=1/1*2+1/2*3+...+1/99*100 

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100<1 (2)

Từ (1) và (2) ta có: A<B<1 <=>A<1

A bé hơn 1 nha bạn