xxxadbbh

1/2.2 < 1/1.2

1/3.3 < 1/2.3

..................

1/100.100 < 1/99.100 

=> <

Ta có: \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)

Vì \(\frac{1}{2^2}<\frac{1}{1.2}\)

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

\(\frac{1}{4^2}<\frac{1}{3.4}\)

.....

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

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

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

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<1\left(đpcm\right)\)

1/2.2 < 1/1.2

1/3.3 < 1/2.3

..................

1/100.100 < 1/99.100 

=> <

rút gọn

1/2.2 + 1/3.3 + 1/4.4 + ... + 1/9.9

> 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10

> 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10

> 1/2 - 1/10

> 5/10 - 1/10

> 2/5 (1)

1/2.2 + 1/3.3 + 1/4.4 + ... + 1/9.9

< 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9

< 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9

< 1 - 1/9

< 8/9 (2)

Từ (1) và (2) => 2/5 < 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/9.9 < 8/9

1/2.2 + 1/3.3 + 1/4.4 +....+ 1/99.99 + 1/100.100

= 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/98.99 + 1/99.100

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99 + 1/99 - 1/100

= 1/1 - 1/100

= 99/100

sai luôn từ bước đầu tiên