Đáp án:
Ta có :
$\frac{1}{4^{2}}$ < $\frac{1}{3.4}$
$\frac{1}{5^{2}}$ < $\frac{1}{4.5}$
......
$\frac{1}{60^{2}}$ < $\frac{1}{59.60}$
=> $\frac{1}{4^{2}}$ + $\frac{1}{5^{2}}$ + .... + $\frac{1}{60^{2}}$ < $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + .... + $\frac{1}{59.60}$
Ta có :
A = $\frac{1}{3.4}$ + $\frac{1}{4.5}$ + .... + $\frac{1}{59.60}$
= $\frac{1}{3}$ - $\frac{1}{4}$ + $\frac{1}{4}$ - $\frac{1}{5}$ + ... + $\frac{1}{59}$ - $\frac{1}{60}$
= $\frac{1}{3}$ - $\frac{1}{60}$ = $\frac{19}{60}$ < $\frac{3}{9}$
=> A + $\frac{1}{3^{2}}$< $\frac{3}{9}$ + $\frac{1}{3^{2}}$ = $\frac{4}{9}$
=> $\frac{1}{3^{2}}$ + $\frac{1}{4^{2}}$ + .... + $\frac{1}{60^{2}}$ < A < $\frac{4}{9}$
=> $\frac{1}{3^{2}}$ + $\frac{1}{4^{2}}$ + .... + $\frac{1}{60^{2}}$ < $\frac{4}{9}$
Giải thích các bước giải:
