집합 원소 개수 합

$\displaystyle n ^{2} + n \leq \frac{1}{2} n + 5$에서 $2 n ^{2} + n - 10 = ( 2 n + 5 ) ( n - 2 ) \leq 0$ 따라서 $n \leq 2$일 때, $\displaystyle n ^{2} + n \leq \frac{1}{2} n + 5$가 성립한다. 따라서 $A _{n} \subset B _{n}$($n = 1$, $2$)이고 $B _{n} \subset A _{n} ( n = 3 , 4 , \cdots )$이다. 따라서 $A _{n} - B _{n} = \emptyset$($n = 1$, $2$)이므로 $a _{n} = 0$이고 $n \geq 3$일 때, $a _{n}$$= n ( A _{n} ) - n ( B _{n} )$$\displaystyle = n ^{2} + n - \left[ \frac{1}{2} n + 5 \right]$이다. 그런데 $n$이 짝수일 때, $\displaystyle \left[ \frac{1}{2} n + 5 \right]$$\displaystyle = \frac{1}{2} n + 5$ $n$이 홀수일 때, $\displaystyle \left[ \frac{1}{2} n + 5 \right]$$\displaystyle = \frac{1}{2} n + 5 - \frac{1}{2}$ 이므로 $\displaystyle \sum\limits _{n=1} ^{20} a _{n}$$\displaystyle = \sum\limits _{n=3} ^{20} \left( n ^{2} + n - \left[ \frac{1}{2} n + 5 \right] \right)$ $\displaystyle = \sum\limits _{n=3} ^{20} \left( n ^{2} + n \right)$$\displaystyle - \sum\limits _{n=3} ^{20} \left( \left[ \frac{1}{2} n + 5 \right] \right)$ 이때, $\displaystyle \sum\limits _{k=1} ^{n} \left( k ^{2} + k \right)$$\displaystyle = \frac{1}{3} n ( n + 1 ) ( n + 2 )$이므로 $\displaystyle \sum\limits _{n=3} ^{20} \left( n ^{2} + n \right)$$\displaystyle = \left\{ \sum\limits _{n=1} ^{20} \left( n ^{2} + n \right) - \sum\limits _{n=1} ^{2} \left( n ^{2} + n \right) \right\}$ $\displaystyle = \frac{1}{3} ( 20 \cdot 21 \cdot 22 - 2 \cdot 3 \cdot 4 )$ $\displaystyle = \frac{1}{3} ( 9240 - 24 )$$= 3072$ 이고 $\displaystyle \sum\limits _{n=3} ^{20} \left( \left[ \frac{1}{2} n + 5 \right] \right)$$\displaystyle = \sum\limits _{n=3} ^{20} \left( \frac{1}{2} n + 5 \right)$$\displaystyle - 9 \cdot \frac{1}{2}$ (∵ $3$~$20$까지 자연수 중 홀수는 $9$개) $\displaystyle = \left( \sum\limits _{n=1} ^{20} \frac{1}{2} n - \sum\limits _{n=1} ^{2} \frac{1}{2} n \right) + 18 \times 5$$\displaystyle - 9 \cdot \frac{1}{2}$ $\displaystyle = \left( 105 - \frac{3}{2} \right) + 90$$\displaystyle - \frac{9}{2}$ $= 189$ 이므로 $\displaystyle \sum\limits _{n=1} ^{20} a _{n}$$= 3072 - 189$$= 2883$