Монгол Бодлогын Сан

Заавар: $A=2+\dfrac{5}{3}+\dfrac{8}{9}+\dfrac{11}{27}+\dfrac{14}{81}+\dots+\dfrac{3n+2}{3^n}+\cdots$ гээд $3A-A$ илэрхийллийг хялбарчил.

Бодолт: $A=\sum\limits_{n=0}^{\infty}\dfrac{3n+2}{3^n}$ тул \begin{align*} 3A-A&=3\sum_{n=0}^{\infty}\dfrac{3n+2}{3^n}-\sum_{n=0}^{\infty}\dfrac{3n+2}{3^n}\\ &=\sum_{n=0}^{\infty}\dfrac{3n+2}{3^{n-1}}-\sum_{n=0}^{\infty}\dfrac{3n+2}{3^n}\\ &=6+\sum_{n=1}^{\infty}\dfrac{3n+2}{3^{n-1}}-\sum_{n=0}^{\infty}\dfrac{3n+2}{3^n}\\ &=6+\sum_{n=1}^{\infty}\dfrac{3(n-1)+5}{3^{n-1}}-\sum_{n=0}^{\infty}\dfrac{3n+2}{3^n}\\ &=6+\sum_{n=0}^{\infty}\dfrac{3n+5}{3^{n}}-\sum_{n=0}^{\infty}\dfrac{3n+2}{3^n}\\ &=6+\sum_{n=0}^{\infty}\dfrac{3}{3^n}=6+\dfrac{3}{1-\frac{1}{3}}=10\dfrac12 \end{align*} тул $A=\dfrac{10\frac12}{2}=5\dfrac14$ байна.