Question

 

Chứng minh S < 2

S = \(\frac{1}{1x1}+\frac{1}{2x2}+\frac{1}{3x3}+\frac{1}{4x4}+...+\frac{1}{10x10}\)

lm được 5 tích 

Answer 1

Ta có: \(S=1+\frac{1}{2x2}+\frac{1}{3x3}+.....+\frac{1}{10x10}\)

Ta có: 1/2x2 < 1/1x2

          1/3x3 < 1/2x3 

          1/4x4 < 1/3x4

         .......................

         1/10x10 < 1/9x10

=> S< 1+1/1x2+1/2x3+1/3x4+.....+1/9x10

=> S<1+(1-1/10)

=> S < 1+9/10

=> S < 19/10 < 2

Vậy S<2 

Answer 2

1/5x5;1/6x6;1/7x7;1/8x8;1/9x9

Answer 3

đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(B=1-\frac{1}{10}<1\) (1)

Mà 1<2 (2)

Ta có:\(S=\frac{1}{1.1}+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{10.10}(3)

Answer 4

S<1+1/1.2+1/2.3+1/3.4+.....+1/9.10

S<1+1-1/2+1/2-1/3+1/3-1/4+.....+1/9-1/10

S<1+1-1/10

S<2-1/10<2

Vậy S<2 (ĐPCM)

Answer 5

S=1/1.1+1/2.2+1/3.3+1/4.4+...+1/10.10

S<1/1+1/1.2+1/2.3+1/3.4+...+1/9.10

S<1+1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

S<2-1/10<2

Chung to S<2
 

Answer 6

Ta có : \(S=1+\frac{1}{2x2}+\frac{1}{3x3}+\frac{1}{4x4}+....+\frac{1}{10x10}<1+\frac{1}{1x2}+\frac{1}{2x3}+.....+\frac{1}{9.10}\)

\(\Rightarrow S<1+\left(1-\frac{1}{10}\right)=1+\frac{9}{10}\)

\(\Rightarrow S<\frac{19}{10}<2\)

Answer 7

Hứa nha=)))

S<2-1/10<2