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

\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}<\frac{3}{10}\)

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}<\frac{3}{40}\)

-> A <\(\frac{1}{8}+\frac{3}{10}+\frac{3}{40}=\frac{20}{40}=\frac{1}{2}\)

A<1/2 nhé!

* S = 3/10+3/11+3/12+3/13+3/14 < 3/10 + 3/10+3/10+3/10+3/10

                                                  < 3/10 x 5

                                                 < 3/2 < 2sư

* S = 3/10+3/11+3/12+3/13+3/14 > 3/15+3/15+3/15+3/15+3/15

                                                > 3/15 x 5

                                               > 1

CHỨNG TỎ ........

                                                 > 1

lon de sua 3/1 thanh 3/10

sua 3/1 thanh 3/10

c ) S = 1.2 + 2.3 + 3.4 + .... + 99.100

=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 99.100.( 101 - 98 )

=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 99.100.101 - 98.99.100

=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 98.99.100 - 98.99.100 ) + 99.100.101

=> 3S = 99.100.101 => S = \(\frac{99.100.101}{3}\)

d ) Ta có \(\frac{1}{2^2}<\frac{1}{2.1}=\frac{1}{1}-\frac{1}{2}\)

               \(\frac{1}{3^2}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

                ..........

                \(\frac{1}{100^2}<\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

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

a,b đề là j bn???????????

phần a và b là tính nhanh