A calculus problem by Priyanshu Mishra

$\begin{aligned} f(x) & = (x^2+8x+20)e^x \\ f'(x) & = (x^2+8x+20)e^x + (2x + 8)e^x = f(x) + \color{#3D99F6} (2x + 8)e^x & \small \color{#3D99F6} \text{Note that } (2x + 8)e^x = f'(x) - f(x) \\ f''(x) & = f'(x) + f'(x) - f(x) + 2e^x = 2f'(x) - f(x) + 2e^x \\ f'''(x) & = 2{\color{#3D99F6}f''(x)} - f'(x) + 2e^x = 2{\color{#3D99F6}(2f'(x) - f(x) + 2e^x)} - f'(x) + 2e^x = 3f'(x) - 2f(x) + 6e^x \\ f^{(4)}(x) & = 4f'(x) - 3f(x) + 12e^x \end{aligned}$

It appears that we can claim that $f^{(n)} = nf'(x) - (n-1)f(x) + n(n-1)e^x$ . Let us prove by induction that the claim is true for $n \ge 1$ . The claim is true for $n=1$ and $n=2$ . Now assuming it is true for $n$ , then

$\begin{aligned} f^{(n+1)} (x) & = nf''(x) - (n-1)f'(x) + n(n-1)e^x \\ & = n(2f'(x) - f(x) + 2e^x) - (n-1)f'(x) + (n^2-n)e^x \\ & = (n+1)f'(x) - nf(x) + (n+1)ne^x \end{aligned}$

Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 1$ . Now we have:

$\begin{aligned} f^{(100)}(x) & = 100f'(x) - 99f(x) + 9900e^x \\ & = 100(x^2+10x+28)e^x - 99(x^2 + 8x+20)e^x + 9900e^x \\ & = (x^2 + 208x+10720)e^x \\ \implies \frac {f^{(100)}(x)}{e^x} & = x^2 + 208x+10720 \\ \frac {f^{(100)}(2018)}{e^{2018}} & = 2018^2 + 208(2018)+10720 \\ & = \boxed{4502788} \end{aligned}$

$let\quad g\left( x \right) ={ x }^{ 2 }+8x+20\\ f\left( x \right) =g\left( x \right) { e }^{ x }\\ f^{ \left( 1 \right) }\left( x \right) =\left( g^{ \left( 1 \right) }\left( x \right) +g\left( x \right) \right) { e }^{ x }\\ f^{ \left( 2 \right) }\left( x \right) =\left( g^{ \left( 2 \right) }\left( x \right) +2g^{ \left( 1 \right) }\left( x \right) +g\left( x \right) \right) { e }^{ x }\\ f^{ \left( 3 \right) }\left( x \right) =\left( g^{ \left( 3 \right) }\left( x \right) +3g^{ \left( 2 \right) }\left( x \right) +3g^{ \left( 1 \right) }\left( x \right) +g\left( x \right) \right) { e }^{ x }\\ \vdots \\ f^{ \left( n \right) }\left( x \right) ={ e }^{ x }\sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } g^{ \left( k \right) }\left( x \right) \\ g^{ \left( 1 \right) }\left( x \right) =2x+8\\ g^{ \left( 2 \right) }\left( x \right) =2\\ for\quad k>2\quad g^{ \left( k \right) }\left( x \right) =0\\ \Rightarrow f^{ \left( 100 \right) }\left( x \right) ={ e }^{ x }\sum _{ k=0 }^{ 100 }{ \left( \begin{matrix} 100 \\ k \end{matrix} \right) } g^{ \left( k \right) }\left( x \right) ={ e }^{ x }\left( \left( \begin{matrix} 100 \\ 0 \end{matrix} \right) g\left( x \right) +\left( \begin{matrix} 100 \\ 1 \end{matrix} \right) g^{ \left( 1 \right) }\left( x \right) +\left( \begin{matrix} 100 \\ 2 \end{matrix} \right) g^{ \left( 2 \right) }\left( x \right) +0+0+\dots \right) \\ \frac { f^{ \left( 100 \right) }\left( 2018 \right) }{ { e }^{ 2018 } } =g\left( 2018 \right) +100g^{ \left( 1 \right) }\left( 2018 \right) +4950g^{ \left( 2 \right) }\left( 2018 \right) =4502788$