TL:
$A =\frac{7}{4}$ $+ \frac{17}{9}$ $+ ...+\frac{4999}{2500}$
$A =( 1+ \frac{3}{4})( 1+ \frac{8}{9})+...+( 1+ \frac{2499}{2500}) ( 49 $ $nhóm )$
$A = 1+ \frac{3}{4} 1+ \frac{8}{9}+...+ 1+ \frac{2499}{2500}$
$A = ( 1 +...+1) + $ $( \frac{3}{4}$ $+\frac{8}{9}+ ...+$ $\frac{2499}{2500})$
$A = 49 + $ $ ( \frac{3}{4}$ $+\frac{8}{9}+ ...+$ $\frac{2499}{2500})$
Đặt B = $ \frac{3}{4}$ $+\frac{8}{9}+ ...+$ $\frac{2499}{2500}$
B = $1-\frac{1}{2^2} + 1 - $ $\frac{1}{3^2} + ... + $ $1 - \frac{1}{50^2}$
$B = 1 + ... + 1 -$ $( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ ... + $ $\frac{1}{50^2})$
$B = 49 - ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ ... + $ $\frac{1}{50^2})$
Thay B ta có:
$A = 49 + B$
$A = 49 + 49 - ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ ... + $ $\frac{1}{50^2})$
$ A = 98 - ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ ... + $ $\frac{1}{50^2})$
$A = 97 + [ 1 - ( \frac{1}{2^2}$ $+ \frac{1}{3^2}+ ... + $ $\frac{1}{50^2}) > 97 $
$=> A > 97$
