등차수열 가중합 시그마

[출제의도] 등차수열을 이용하여 수열의 합을 구한다. 등차수열 $\left\{ a _{n} \right\}$의 공차를 $d$라 하면 $a _{n} = 1 + \left( n - 1 \right) d$이므로 $b _{n}$$\displaystyle = \sum\limits _{k = 1} ^{n} k \left\{ 1 + \left( k - 1 \right) d \right\}$ $\displaystyle = \sum\limits _{k = 1} ^{n} \left\{ dk ^{2} + \left( 1 - d \right) k \right\}$ $\displaystyle = d \sum\limits _{k = 1} ^{n} k ^{2} + \left( 1 - d \right) \sum\limits _{k = 1} ^{n} k$ $\displaystyle = \frac{dn \left( n + 1 \right) \left( 2 n + 1 \right)}{6} + \frac{\left( 1 - d \right) n \left( n + 1 \right)}{2}$ $\displaystyle = \frac{n \left( n + 1 \right)}{6} \times \left\{ d \left( 2 n + 1 \right) + 3 \left( 1 - d \right) \right\}$ $\displaystyle = \frac{n \left( n + 1 \right)}{6} \times \left( 2 dn - 2 d + 3 \right)$ $b _{10} = 715$이므로 $\displaystyle \frac{10 \times 11}{6} \times \left( 20 d - 2 d + 3 \right) = 715$ $18 d + 3 = 39$ $18 d = 36$ $d = 2$ $b _{n}$$\displaystyle = \frac{n \left( n + 1 \right)}{6} \times \left( 2 \times 2 \times n - 2 \times 2 + 3 \right)$ $\displaystyle = \frac{n \left( n + 1 \right)}{6} \times \left( 4 n - 1 \right)$ $\displaystyle \frac{b _{n}}{n \left( n + 1 \right)} = \frac{1}{6} \left( 4 n - 1 \right)$ 따라서 $\displaystyle \sum\limits _{n = 1} ^{10} \frac{b _{n}}{n \left( n + 1 \right)}$$\displaystyle = \sum\limits _{n = 1} ^{10} \frac{1}{6} \left( 4 n - 1 \right)$$\displaystyle = \frac{1}{6} \left( 220 - 10 \right)$$= 35$ [다른 풀이] 등차수열 $\left\{ a _{n} \right\}$의 공차를 $d$라 하면 $a _{n} = 1 + \left( n - 1 \right) d$이므로 $b _{n}$$\displaystyle = \sum\limits _{k = 1} ^{n} k \left\{ 1 + \left( k - 1 \right) d \right\}$ $\displaystyle = \sum\limits _{k = 1} ^{n} \left\{ dk ^{2} + \left( 1 - d \right) k \right\}$ $\displaystyle = d \sum\limits _{k = 1} ^{n} k ^{2} + \left( 1 - d \right) \sum\limits _{k = 1} ^{n} k$ $\displaystyle = \frac{dn \left( n + 1 \right) \left( 2 n + 1 \right)}{6} + \frac{\left( 1 - d \right) n \left( n + 1 \right)}{2}$ $\displaystyle = \frac{n \left( n + 1 \right)}{6} \times \left\{ d \left( 2 n + 1 \right) + 3 \left( 1 - d \right) \right\}$ $\displaystyle = \frac{n \left( n + 1 \right)}{6} \times \left( 2 dn - 2 d + 3 \right)$ $\displaystyle \frac{b _{n}}{n \left( n + 1 \right)}$$\displaystyle = \frac{3 - 2 d + 2 dn}{6}$$\displaystyle = \frac{1}{2} + \left( n - 1 \right) \frac{d}{3}$ 이므로 수열 $\displaystyle \left\{ \frac{b _{n}}{n \left( n + 1 \right)} \right\}$은 첫째항이 $\displaystyle \frac{1}{2}$이고 공차가 $\displaystyle \frac{d}{3}$인 등차수열이다. $b _{10} = 715$이므로 수열 $\displaystyle \left\{ \frac{b _{n}}{n \left( n + 1 \right)} \right\}$의 제$10$항은 $\displaystyle \frac{b _{10}}{10 \times 11} = \frac{715}{10 \times 11} = \frac{13}{2}$ 따라서 $\displaystyle \sum\limits _{n = 1} ^{10} \frac{b _{n}}{n \left( n + 1 \right)}$$\displaystyle = \frac{10 \times \left( \frac{1}{2} + \frac{13}{2} \right)}{2}$$\displaystyle = \frac{10 \times 7}{2}$$= 35$