Will you integrate it 2015 times? - IV

Let $P(x) = \displaystyle \sum_{i = 0 } ^{2015} a_{i} x^{i} = a_{0} + a_{1} x + a_{2} x^{2} + \cdots + a_{2015} x^{2015}$

After differentiating it $2015$ times, we get $Q(x) = 2015! \times a_{2015}$

$\begin{aligned} &P(x) = Q(x) \\ \implies &a_{0} + a_{1} x + a_{2} x^{2} + \cdots + a_{2015} x^{2015} = 2015! \times a_{2015} \\ \implies & (a_{0} - 2015! \times a_{2015})+ a_{1} x + a_{2} x^{2} + \cdots + a_{2015} x^{2015} = 0\\ \end{aligned}$

This is just a polynomial of degree 2015 with real coefficients, and can have any number of distinct roots from $1$ to $2015$ . This means $n \in \{ 1, 2, 3, \ldots, 2015\}$ .

Therefore sum of all possible values of $n = 1 + 2 + 3 +\cdots + 2015$ $= \dfrac{ 2015 \times 2016 } { 2} = \boxed{ 2031120 }$ $_\square$