점화식 빈칸

$\displaystyle a _{n+1} - 2 \sum\limits _{k=1} ^{n} \frac{a _{k}}{k} = 2 ^{n+1} ( n ^{2} + n + 2 )$ ($n \geq 1$) ⋯⋯ ㉠ ㉠의 $n$대신 $n - 1$을 대입하면 $\displaystyle a _{n} - 2 \sum\limits _{k=1} ^{n-1} \frac{a _{k}}{k} = 2 ^{n} ( n ^{2} - n + 2 )$ ⋯⋯ ㉡ 이다. ㉠$-$㉡에 의하여 $\displaystyle a _{n+1} - a _{n} - \frac{2}{n} a _{n} = \square {2 ^{n} ( n ^{2} + 3 n + 2 )}$ ( $n \geq 2$) 이므로 $\displaystyle a _{n+1} - \frac{n + 2}{n} a _{n} =$$2 ^{n} ( n + 1 ) ( n + 2 )$ 즉, $\displaystyle \frac{a _{n+1}}{( n + 1 ) ( n + 2 )} - \frac{a _{n}}{n ( n + 1 )} = 2 ^{n}$ 이다. $\displaystyle b _{n} = \frac{a _{n}}{n ( n + 1 )}$이라 하면 $b _{n+1} - b _{n} = \square {2 ^{n}}$ ($n \geq 2$) 이고 $b _{2} = 0$이므로 $b _{n}$$\displaystyle = b _{2} + \sum\limits _{k=2} ^{n-1} ( b _{k+1} - b _{k} )$$\displaystyle = \sum\limits _{k=2} ^{n-1} 2 ^{k}$$\displaystyle = \frac{4 ( 2 ^{n-2} - 1 )}{2 - 1}$ $= \square {2 ^{n} - 4}$ 이다. ∴ $f ( n ) = 2 ^{n} ( n ^{2} + 3 n + 2 )$ $g ( n ) = 2 ^{n}$ $h ( n ) = 2 ^{n} - 4$ ∴ $\displaystyle \frac{f ( 4 )}{g ( 5 )} + h ( 6 )$$\displaystyle = \frac{2 ^{4} \cdot 30}{2 ^{5}} + ( 2 ^{6} - 4 )$$= 75$