Ta gọi tổng cần tìm là S, vậy ta có:
S=\frac{1}{6}.(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{380}
S= \frac{1}{6}.(\frac{1}{1.2}+\frac{1}{2.3}\frac{1}{3.4}+...+\frac{1}{19.20}
S= \frac{1}{6}.(1-\frac{1}{20}
S= \frac{1}{6}.\frac{19}{20}
S=\frac{19}{120}
Dấu (.) là nhân (x) $\dfrac{1}{2.6}$+$\dfrac{1}{4.9}$+$\dfrac{1}{6.12}$+...+$\dfrac{1}{36.57}$+$\dfrac{1}{39.60}$
\Rightarrow $\dfrac{1}{2.6}$+$\dfrac{1}{6.6}$+$\dfrac{1}{6.12}$+...+$\dfrac{1}{6.6.57}$+$\dfrac{1}{39.6.10}$
\Rightarrow $\dfrac{1}{6}$.{$\dfrac{1}{2}$+$\dfrac{1}{6}$+$\dfrac{1}{12}$+...+$\dfrac{1}{6.57}$+$\dfrac{1}{39.10}$}
\Rightarrow Xong tu giải