등차수열 합 빈칸 추론

수열 $\left\{ a _{n} \right\}$이 등차수열이므로 자연수 $n$에 대하여 $\displaystyle S _{n} = \frac{n \left( a _{1} + a _{n} \right)}{2}$ $\cdots$$\cdots$ ㉠ 수열 $\left\{ a _{n} \right\}$의 공차를 $d$라 하자. (ⅰ) $n = 2 k - 1$ ($k$는 자연수)일 때, ㉠에 의하여 $b _{2 k - 1}$$\displaystyle = 2 \times \left\{ \frac{\left( 2 k - 1 \right) \left( a _{1} + a _{2 k - 1} \right)}{2} \right.$$\displaystyle \left. \begin{array}{l} - \left( 2 k - 1 \right) a _{2 k - 1} \begin{matrix} \\ \end{matrix} \end{array} \right\}$ $= \left( 2 k - 1 \right) \left( a _{1} + a _{2 k - 1} \right) - 2 \left( 2 k - 1 \right) a _{2 k - 1}$ $= \left( \square {2 k - 1} \right) \times \left( a _{1} - a _{2 k - 1} \right)$ (ⅱ) $n = 2 k$ ($k$는 자연수)일 때, $b _{2 k} = \left( 2 k - 1 \right) a _{2 k}$ $b _{2} = \left( 2 - 1 \right) a _{2} = a _{2} = 2$ (ⅰ), (ⅱ)에 의하여 $b _{2 k - 1} + b _{2 k}$ $= \left( \square {2 k - 1} \right) \times \left( a _{1} - a _{2 k - 1} \right) + \left( 2 k - 1 \right) a _{2 k}$ $= \left( 2 k - 1 \right) \times \left\{ a _{1} + \left( a _{2 k} - a _{2 k - 1} \right) \right\}$ $= \left( 2 k - 1 \right) \times \left( a _{1} + d \right)$ $= \left( 2 k - 1 \right) \times a _{2}$ $= \square {4 k - 2}$ $\cdots$$\cdots$ ㉡ 이다. ㉡에 의하여 $\displaystyle \sum\limits _{n = 1} ^{20} b _{n}$$= \left( b _{1} + b _{2} \right) + \left( b _{3} + b _{4} \right) + \cdots + \left( b _{19} + b _{20} \right)$ $\displaystyle = \sum\limits _{k = 1} ^{10} \left( b _{2 k - 1} + b _{2 k} \right)$ $\displaystyle = \sum\limits _{k = 1} ^{10} \left( 4 k - 2 \right)$ $\displaystyle = 4 \times \frac{10 \times 11}{2} - 2 \times 10$ $= \square {200}$ 이다. $f \left( k \right) = 2 k - 1$, $g \left( k \right) = 4 k - 2$, $p = 200$ 따라서 $f \left( 2 \right) + g \left( 3 \right) + p = 3 + 10 + 200 = 213$