Монгол Бодлогын Сан

Заавар: $$\dfrac{1}{(n+\alpha)(n+\beta)}=\dfrac{1}{\beta-\alpha}\left(\dfrac{1}{n+\alpha}-\dfrac{1}{n+\beta}\right)$$ ашигла.

Бодолт: \begin{align*} \dfrac{1}{n(n+1)(n+2)}&=\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\dfrac{1}{n+2}\\ &=\dfrac{1}{n(n+2)}-\dfrac{1}{(n+1)(n+2)}\\ &=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-\dfrac{1}{n+1}+\dfrac{1}{n+2}\\ &=\dfrac12\cdot\dfrac{1}{n}-\dfrac{3}{2}\cdot\dfrac{1}{n+1}+\dfrac{1}{n+1} \end{align*} байна.