Đáp án:
c) $\frac{2²}{3.5}+\frac{2²}{5.7}+\frac{2²}{7.9}+...+\frac{2²}{97.99}$
$=2.(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99})$
$=2.(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99})$
$=2.(\frac{1}{3}-\frac{1}{99})=2.\frac{32}{99}=\frac{64}{99}$
#NOCOPY