Question

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

Answer 1

1/2.2 < 1/1.2

1/3.3 < 1/2.3

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

1/100.100 < 1/99.100 

=> <

Answer 2

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 \(\frac{1}{2^2}<\frac{1}{1.2}\)

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

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

.....

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

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

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

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

Answer 3

1/2.2 < 1/1.2

1/3.3 < 1/2.3

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

1/100.100 < 1/99.100 

=> <