B1:Chứng Minh rằng A=1/”`3^2″`+1/”`4^2″`+1/”`5^2″`+…+1/”`50^2″` >1/4

Giải thích các bước giải:

$3^2 <3.4$
$4^2<4.5$
…
$50^2<50.51$
ta có $A= \dfrac{1}{3^2}+\dfrac{1}{4^2}+..+\dfrac{1}{50^2} > \dfrac{1}{3.4}+ \dfrac{1}{4.5}+..+\dfrac{1}{50.51}$
$=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+..+\dfrac{1}{50}-\dfrac{1}{51}$
$=\dfrac{1}{3}-\dfrac{1}{51}>\dfrac{1}{3}-\dfrac{1}{12}=\dfrac{1}{4}$