Zdroj TeX:

\sum_{i=1}^{n}{ \frac{1}{(3i-2)}. \frac{1}{(3i+1)} } =\sum_{i=1}^{n}\frac{ \frac{1}{3} }{(3i-2)}-\frac{ \frac{1}{3} }{(3i+1)}=