Derivative Magic II

$\large{\begin{aligned}f(x)&=\displaystyle\sum_{k=1}^{n} \dfrac{x^{k+1}}{k+1}\\f'(x)&=\displaystyle\sum_{k=1}^{n} \dfrac{\cancel{(k+1)}x^{k}}{\cancel{k+1}}\\f^2(x)&=\displaystyle\sum_{k=1}^{n} kx^{k-1}\\f^3(x)&=\displaystyle\sum_{k=1}^{n}k( k-1)x^{k-2}\\&\cdots\\&\cdots\\ f^n~(x)&=\displaystyle\sum_{k=1}^{n}(k)(k-1)\cdots(k-(n-2))x^{(k-(n-1))}\end{aligned}}$

Since first $(n-2)$ terms of the summation will be zero we only have to calculate last two terms i.e:

$=\displaystyle\sum_{k=(n-1)}^{n}((k)(k-1)\cdots(k-(n-2))x^{(k-(n-1))}$

$\large{ =(n-1)!+xn!\\=\boxed{(n-1)!(1+nx)}}$

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In the given question, $n=2016, x=100$ . Putting we get $2015!\times (2016001)$ . Hence answer $=2015+2016001=\boxed{203616}$ .

We know that $f(x)=\sum_{k=1}^{2016}\frac{x^{k+1}}{k+1}=\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+...+\frac{x^{2017}}{2017}$ $\Rightarrow f^{1}(x)=x+x^{2}+x^{3}+...+x^{2016}$ $\Rightarrow f^{2016}(x)=\prod_{k=1}^{2015}k+(\prod_{k=1}^{2016}k)x$ Now, we get $f^{2016}(100)=\prod_{k=1}^{2015}k+(\prod_{k=1}^{2016}k)(100)$ $=(\prod_{k=1}^{2015}k)\times[1+2016(100)]$ $=(\prod_{k=1}^{2015}k)\times201601$ The answer is now in the desired form $\therefore 2015+201601=\boxed{203616}$ .