Mk nghĩ là thê này nè , k biêt co đung k:
Ta co \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\) (1)
\(.......\)
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
Từ (1) \(=>B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)\(=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
\(=>B< 1\left(đpcm\right)\)
tick cho mk nha
Ta có: B = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{100^2}\)< \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{99.100}\)
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{100^2}\) < 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\).
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{100^2}\) < 1 - \(\dfrac{1}{100}\)< 1.
Vậy B < 1