Giải thích các bước giải:
Bạn tham khảo:
a) $(1+3+5+...+2007+2009+2011)(125125.127-127127.125)$
$=(1+3+5+...+2007+2009+2011)(125.1001.127-127127.125)$
$=(1+3+5+...+2007+2009+2011)(125.127127-127127.125)$
$=(1+3+5+...+2007+2009+2011).0$
$=0$
b) $\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}$
$=\frac{1}{1}.\frac{1}{3}+\frac{1}{3}.\frac{1}{5}+\frac{1}{5}.\frac{1}{7}+\frac{1}{7}.\frac{1}{9}+\frac{1}{9}.\frac{1}{11}+\frac{1}{11}.\frac{1}{13}+\frac{1}{13}.\frac{1}{15}$
$=\frac{1}{2}(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15})$
$=\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15})$
$=\frac{1}{2}(1-\frac{1}{15})$
$=\frac{1}{2}.\frac{14}{15}$
$=\frac{14}{30}=\frac{7}{15}$
