Question

Chứng tỏ rằng: S = \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}<1\)

Giúp mình với!

Answer 1

S = 1/2+1/2.3+1/3.4 +... +1/9/10

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

S =1-10

S =9/10

Do 9/10<1

=>S<1

Answer 2

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

  =1-(1/2-1/2)-(1/3-1/3)-(1/4-1/4)-...-(1/9-1/9)-1/10

  =1-1/10<1

Vậy S<1

Answer 3

Ta có 

\(S=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\)

\(S=1-\frac{1}{10}<1\)(ĐFCM)

Answer 4

từ đó suy ra 

s,

Answer 5

s=1/1.2+1/2.3+1/3.4+.........+1/9.10

s=1-1/2+1/2-1/3+.............+1/9-1/10

s=1-1/10<1

=>s<1

h mk nhe