Question

CMR:\(\frac{1}{6}\)<\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+.....+\(\frac{1}{100^2}\)<\(\frac{1}{4}\)

 

Answer 1

:\(\frac{1}{6}\)<\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+.....+\(\frac{1}{100^2}\)<\(\frac{1}{4}\)

=\(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+\(\frac{1}{7^2}\)+.....+\(\frac{1}{100^2}\)<\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+.....+\(\frac{1}{99.100}\)

=\(\frac{1}{4}\)-\(\frac{1}{100}\)=\(\frac{24}{100}\)<\(\frac{25}{100}\)=\(\frac{1}{4}\)>\(\frac{20}{100}\)=\(\frac{1}{5}\)>\(\frac{1}{6}\)

Vậy \(\frac{1}{6}\)<\(\frac{1}{5}\)

Answer 2

:$\frac{1}{6}$16 <$\frac{1}{5^2}$1‍52 +$\frac{1}{6^2}$162 +$\frac{1}{7^2}$‍172 +.....+$\frac{1}{100^2}$11002 <$\frac{1}{4}$14 

=>$\frac{1}{5^2}$152 +$\frac{1}{6^2}$162 +$\frac{1}{7^2}$172 +.....+ 

=> bạn biết làm rồi nên thôi 

=> nói thật ra là bí

Answer 3

1/6 < 1/5< 1/4