도함수 점화식 다항식의 상수항

$f _{1} ( x ) = x ^{2}$에서 $f' _{1} ( x ) = 2 x$ $f _{n+1} ( x ) = f _{n} ( x ) + f' _{n} ( x )$에서 $f _{2} ( x ) = f _{1} ( x ) + f' _{1} ( x ) = x ^{2} + 2 x$ 즉, $f _{n} ( x )$는 $x ^{2}$인 계수가 $1$인 $2$차식이므로 $f _{n} ( x ) = x ^{2} + a _{n} x + b _{n}$으로 놓을 수 있다. $f _{n+1} ( x ) = x ^{2} + a _{n+1} x + b _{n+1}$로 놓으면 $f _{n+1} ( x ) - f _{n} ( x ) = f' _{n} ( x )$에서 $x ^{2} + a _{n+1} x + b _{n+1} - ( x ^{2} + a _{n} x + b _{n} )$$= 2 x + a _{n}$ $( a _{n+1} - a _{n} ) x + b _{n+1} - b _{n} = 2 x + a _{n}$ $\therefore a _{n + 1} - a _{n} = 2 , b _{n + 1} - b _{n} = a _{n}$ $a _{n+1} - a _{n} = 2$에서 $a _{n} = a _{1} + 2 ( n - 1 )$ $a _{1} = 0$이므로$a _{n} = 2 ( n - 1 )$ $b _{n+1} - b _{n} = 2 ( n - 1 )$에서$\displaystyle b _{n} = b _{1} + \sum\limits _{k=1} ^{n-1} 2 ( k - 1 )$ $\displaystyle \therefore b _{25} = b _{1} + \sum\limits _{k = 1} ^{24} 2 ( k - 1 )$$\displaystyle = 0 + 2 \left( \frac{24 \cdot 25}{2} \right) - 2 \cdot 24$$= 552$