정수 수열 합의 최대최소

$3 a _{n} ^{2} + 2 na _{n} - 8 n ^{2} = 0$에서 $\left( 3 a _{n} - 4 n \right) \left( a _{n} + 2 n \right) = 0$ $\displaystyle a _{n} = \frac{4}{3} n$ 또는 $a _{n} = - 2 n$ 수열 $\left\{ a _{n} \right\}$의 모든 항이 정수이므로 자연수 $k$에 대하여 $\displaystyle a _{n} = \begin{cases} - 6 k + 4 & & \left( n = 3 k - 2 \right) \\ - 6 k + 2 & & \left( n = 3 k - 1 \right) \\ - 6 k \text{또는} 4 k & & \left( n = 3 k \right) \end{cases}$ $- 6 k < 0$, $4 k > 0$이므로 $1 \leq k \leq 10$인 모든 자연수 $k$에 대하여 $a _{3 k} = 4 k$일 때, $\displaystyle \sum\limits _{n = 1} ^{30} a _{n}$은 최댓값 $M$을 갖고 $a _{3 k} = - 6 k$일 때, $\displaystyle \sum\limits _{n = 1} ^{30} a _{n}$은 최솟값 $m$을 갖는다. $\displaystyle M = \sum\limits _{n = 1} ^{30} a _{n} = \sum\limits _{k = 1} ^{10} a _{3 k - 2} + \sum\limits _{k = 1} ^{10} a _{3 k - 1} + \sum\limits _{k = 1} ^{10} 4 k$이고, $\displaystyle m = \sum\limits _{n = 1} ^{30} a _{n} = \sum\limits _{k = 1} ^{10} a _{3 k - 2} + \sum\limits _{k = 1} ^{10} a _{3 k - 1} + \sum\limits _{k = 1} ^{10} \left( - 6 k \right)$이므로 $M - m$$\displaystyle = \left( \sum\limits _{k = 1} ^{10} a _{3 k - 2} + \sum\limits _{k = 1} ^{10} a _{3 k - 1} + \sum\limits _{k = 1} ^{10} 4 k \right)$ $\displaystyle - \left\{ \sum\limits _{k = 1} ^{10} a _{3 k - 2} + \sum\limits _{k = 1} ^{10} a _{3 k - 1} + \sum\limits _{k = 1} ^{10} \left( - 6 k \right) \right\}$ $\displaystyle = \sum\limits _{k = 1} ^{10} 4 k - \sum\limits _{k = 1} ^{10} \left( - 6 k \right)$ $\displaystyle = \sum\limits _{k = 1} ^{10} \left\{ 4 k - \left( - 6 k \right) \right\}$ $\displaystyle = \sum\limits _{k = 1} ^{10} 10 k = 10 \sum\limits _{k = 1} ^{10} k$ $\displaystyle = 10 \times \frac{10 \times 11}{2} = 550$