점화식 일반항 빈칸

[출제의도] 수열의 귀납적 정의를 이용하여 일반항 추론하기 $\displaystyle b _{n} = \frac{a _{n}}{n + 1}$이라 놓으면 $a _{n} = ( n + 1 ) b _{n}$이므로 $\left( n + 3 \right) b _{n + 2} = \square {( n + 2 )} b _{n + 1} + b _{n}$ $\left( n + 3 \right) \left( b _{n + 2} - b _{n + 1} \right) = - \left( b _{n + 1} - b _{n} \right)$$\cdots$$\cdots$ (★) 식 (★)에 $n = 1 , 2 , \cdots , m - 1$ $\left( m \geq 2 \right)$를 대입하면 $4 \left( b _{3} - b _{2} \right) = - \left( b _{2} - b _{1} \right)$ $5 \left( b _{4} - b _{3} \right) = - \left( b _{3} - b _{2} \right)$ $\vdots$ $\left( m + 2 \right) \left( b _{m + 1} - b _{m} \right) = - \left( b _{m} - b _{m - 1} \right)$ 좌변과 우변을 각각 곱하여 정리하면, $\displaystyle b _{m + 1} - b _{m} = \left( - \frac{1}{4} \right) \left( - \frac{1}{5} \right) \cdots \left( - \frac{1}{m + 2} \right) ( b _{2} - b _{1} )$ $\displaystyle \left. b _{n} = b _{1} + \sum\limits _{k = 1} ^{n - 1} ( b _{k + 1} - b _{k} ) \right.$ $( n \geq 2 )$ $\displaystyle \therefore a _{n} = \left( n + 1 \right) \left( \frac{1}{2} + \sum\limits _{k = 1} ^{n - 1} \square {\frac{( - 1 ) ^{k - 1}}{( k + 2 ) !}} \right)$ $( n \geq 2 )$ $f ( n ) = n + 2$, $\displaystyle g ( k ) = \frac{( - 1 ) ^{k - 1}}{( k + 2 ) !}$ $\displaystyle \therefore f ( 1 ) g ( 3 ) = \frac{1}{40}$