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Define $f(x)=\displaystyle\sum_{k=1}^{2016} \dfrac{x^{k+1}}{k+1}$ . If $f^{(2016)} (100)$ can be evaluated as $a\times b!$ where $b$ is maximum, find $a+b$ .

Bonus: If $f(x)=\displaystyle\sum_{k=1}^{n} \dfrac{x^{k+1}}{k+1}$ , find $f^{(n)} (x)$ .

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