Question

Chứng tỏ rằng :

a) 1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/99.100 < 1

b) 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2

Answer 1

a) 1/1.2 + 1/2.3 + 1/3.4 + ....... + 1/99.100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/99 - 1/100

= 1 - 1/100

= 99/100 < 1 nên 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100 < 1 (ĐPCM)

Answer 2

a)1-1/2+1/2-1/3+1/3-1/4+......+1/99-1/100

1-1/100=99/100<1

cho mk nha ^^

Answer 3

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

Vì \(\frac{1}{100}>0\Rightarrow1-\frac{1}{100}< 1\)

\(\frac{\Rightarrow1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}< 1\)

Answer 4

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)

= \(\frac{1}{1}.\frac{1}{2}-\frac{1}{2}.\frac{1}{3}-\frac{1}{3}.\frac{1}{4}-......-\frac{1}{99}.\frac{1}{100}\)( gạch các số đối nhau thì ta còn lại )

= \(\frac{1}{1}-\frac{1}{100}\)

= \(\frac{100}{100}-\frac{1}{100}\)

= \(\frac{99}{100}\)