Монгол Бодлогын Сан

Заавар:$a_n=b_{n+1}-b_{n}$, $n\in\mathbb Z$ бол $a_1+a_2+\dots+a_n=b_{n+1}-b_{1}$ байдаг. $$\dfrac{1}{n(n+2)}=\dfrac12\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)$$ болохыг ашиглан бод.

Бодолт:\begin{align*} \tfrac{1}{1\cdot 3}+\tfrac{1}{3\cdot 5}+\dots+\tfrac{1}{99\cdot 101}&=\tfrac12\left(\tfrac1{1}-\tfrac1{3}+\tfrac13-\tfrac15+\dots+\tfrac1{99}-\tfrac1{101}\right)\\ &=\tfrac12\left(\tfrac11-\tfrac1{101}\right)=\tfrac12\cdot\tfrac{100}{101}=\tfrac{50}{101} \end{align*}