합 점화식 빈칸

$\displaystyle \frac{{S} _{n + 1}}{n + 1} = \sum\limits _{k = 1} ^{n} {S} _{k} ( n \geq 1 ) \cdots \cdots ( * )$ 에 의하여 $\displaystyle \frac{{S} _{n}}{n} = \sum\limits _{k = 1} ^{n - 1} {S} _{k} ( n \geq 2 ) \cdots \cdots \text{㉠}$ (*)에서 ㉠을 빼면 $\displaystyle \frac{S _{n + 1}}{n + 1} - \frac{S _{n}}{n} =$$\displaystyle \sum\limits _{k=1} ^{n} S _{k} - \sum\limits _{n-1} ^{n} S _{k}$ $\displaystyle \frac{S _{n+1}}{n + 1} - \frac{S _{n}}{n} = S _{n}$ $( n \geq 2 )$ $\displaystyle \frac{S _{n+1}}{n + 1} = \frac{n + 1}{n} S _{n}$$\displaystyle = \frac{( n + 1 ) S _{n}}{n}$ $\therefore$ $\displaystyle \frac{S _{n+1}}{S _{n}} = \square \frac{( n + 1 ) ^{2}}{n}$ $( n \geq 2 )$ ㉠으로부터 $S _{2} = 2$이고, $\displaystyle {S} _{n} = \frac{{S} _{n}}{{S} _{n-1}} \times \frac{{S} _{n-1}}{{S} _{n-2}} \times \cdots \times \frac{{S} _{3}}{{S} _{2}} \times {S} _{2} ( n \geq 3 )$ 이므로 $S _{n}$$\displaystyle = \frac{n ^{2}}{n - 1} \times \frac{( n - 1 ) ^{2}}{( n - 2 )} \times \cdots \times \frac{3 ^{2}}{2} \times 2$ $=$$\displaystyle n \cdot n ! \times \frac{1}{2 \times 1}$ $\displaystyle = n ! \times \square {\frac{n}{2}}$ $( n \geq 3 )$이다. 그러므로 $a _{n}$은 $\displaystyle a _{n} = {\begin{cases} 1 & ( n = 1 , 2 ) \\ \frac{n ^{2} - n + 1}{2} \times ( n - 1 ) ! & ( n \geq 3 ) \end{cases}}$이다. $\therefore$ $\displaystyle f ( n ) = ( n + 1 ) ^{2} , g ( n ) = \frac{n}{2}$ $\displaystyle \therefore f ( 4 ) \times g ( 20 ) = 5 ^{2} \times \frac{20}{2} = 250$