Being brave 1

Well Akshat, first of all thanks for sharing such a delightful problem with the community.

Now here is the solution,

First we will construct the polynomial that is given in the question.

$P(x) = (1)^{2}x^{2015}+(2)^{2}x^{2014}+...+(2015)^{2}x+(2016)^{2}$

$\Rightarrow P(x) = x^{2015}+4x^{2014}+...+4060225x+4064256$

Note that $x_{1},x_{2},x_{3},...,x_{2015}$ are the roots of the above polynomial,

Now the expression whose value we need to calculate,

$\displaystyle\sum_{j=0}^{2015}\displaystyle\sum_{i=0}^{2015}x_{i}^{j}$

$\Rightarrow \displaystyle\sum_{i=0}^{2015}x_{i}^{0}+\displaystyle\sum_{i=0}^{2015}x_{i}^{1}+\displaystyle\sum_{i=0}^{2015}x_{i}^{2}+...+\displaystyle\sum_{i=0}^{2015}x_{i}^{2015}$

In Newton's Sum Method , we write $\displaystyle\sum_{i=0}^{2015}x_{i}^{j}=S_{j}$

So,

$S_{0}+S_{1}+S_{2}+...+S_{2015}$

Now applying the Newton's Sum Identities,

$(1)S_{1}+(1)(4)=0 \Rightarrow S_{1}=-4$

$(1)S_{2}+(4)(-4)+(2)(9)=0 \Rightarrow S_{2}=-2$

$(1)S_{3}+(4)(-2)+(9)(-4)+(3)(16)=0 \Rightarrow S_{3}=-4$

$(1)S_{4}+(4)(-4)+(9)(-2)+(16)(-4)+(4)(25)=0 \Rightarrow S_{4}=-2$

$\vdots$

$S_{2014}=-2$

$S_{2015}=-4$

Now just adding all the valuse,

$\Rightarrow 2015+1008(-4)+1007(-2)$

$\Rightarrow -4031$