Ta có:\(\frac{1}{5.6}\)<\(\frac{1}{5^2}<\frac{1}{4.5}\)
\(\frac{1}{6.7}\) \(\frac{1}{6^2}<\frac{1}{5.6}\)....
\(\frac{1}{100,101}<\frac{1}{100^2}<\frac{1}{99.100}\)
=>\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}<\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
<=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}<\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{5}-\frac{1}{101}
Đặt :
A=1/5^2+1/6^2+...+1/100^2
Ta có:
A<1/4.5+1/5.6+...+1/99.100=1/4-1/5+1/5-1/6+...+1/99-1/100=1/4-1/100<1/4
Đúng thì k nha!
Ta có:
A>1/5.6+1/6.7+...+1/100.101=1/5-1/6+1/6-1/7+....+1/100+1/101>1/6
