합 조건 수열의 일반항 빈칸

$\displaystyle a _{n + 1} = n \cdot 2 ^{n} + \sum\limits _{k = 1} ^{n} \frac{a _{k}}{k} \cdots \cdots \text{㉠}$ $\displaystyle a _{n} = ( n - 1 ) \cdot 2 ^{n - 1} + \sum\limits _{k = 1} ^{n - 1} \frac{a _{k}}{k} \cdots \cdots \text{㉡}$ $\text{㉠} - \text{㉡}$ 하면 $\displaystyle a _{n + 1} - a _{n} = \square {( n + 1 ) \cdot 2 ^{n - 1}} + \frac{a _{n}}{n}$ $\displaystyle a _{n + 1} = \frac{( n + 1 ) a _{n}}{n} + \square {( n + 1 ) \cdot 2 ^{n - 1}}$ $\displaystyle \frac{a _{n+1}}{n + 1} = \frac{a _{n}}{n} + 2 ^{n-1}$ $\displaystyle b _{n} = \frac{a _{n}}{n}$이라 하면 $\displaystyle b _{n + 1} = b _{n} + \frac{\square {( n + 1 ) \cdot 2 ^{n - 1}}}{n + 1} ( n \geq 2 )$ $b _{2} = 3$이므로 $\displaystyle b _{n} = 3 + \sum\limits _{k=1} ^{n-2} 2 ^{k}$$\displaystyle = 3 + \frac{2 ( 2 ^{n-2} - 1 )}{2 - 1}$$= \square {2 ^{n - 1} + 1} ( n \geq 2 )$ $\displaystyle \therefore a _{n} = {\begin{cases} 4 ( n = 1 ) \\ n ( 2 ^{n - 1} + 1 ) ( n \geq 2 ) \end{cases}}$ $\therefore f ( n ) = ( n + 1 ) \cdot 2 ^{n - 1}$, $g ( n ) = 2 ^{n-1} + 1$ 따라서 $f ( 4 ) + g ( 7 ) = 5 \cdot 2 ^{3} + 2 ^{6} + 1$$= 40 + 64 + 1$$= 105$