$f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$ (From OMITS Indonesian $2015$)
Known that $f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$, if $x_1, x_2, x_3,..., x_{2015}$ is the roots of $f(x)$, Find :

$\sum_{n=0}^{2014}$ $\sum_{k=1}^{2015}$ $(x_k)^n$

$a. -2016$
$b. -2015$
$c. -2014$
$d. -2013$
$e. -2012$
What should i do? and where i can learn more about it?
 A: First, we note that $f(x)-f(x)/x=\frac{-2016}{x}+\sum_{k=0}^{2015} x^k$  Dividing by $x$ and rearranging, we get
$$\sum_{k=0}^{2014} x^k=f(x)/x-f(x)/x^2+2016/x^2 -1/x$$
Evaluating at the roots of $f(x)$ and summing yields
$$\sum_i\sum_{k=0}^{2014} x_i^k=\sum_i 2016/x_i^2 -1/x_i.$$
This reduces the problem to finding the sum of the reciprocals of the roots, and finding the sum of the squares of the reciprocals of the roots.  This is an exercise in rewriting symmetric functions in terms of elementary symmetric polynomials to rewrite the expression in terms of the coefficients of $f(x)$.
A: $f(x) = x^{2015}+2x^{2014}+3x^{2013}+...+2015x+2016$.
Take this as a particular example of the class of polynomials $$f_d(x)=x^d+2x^{d-1}+3x^{d-2}+\cdots+dx+(d+1)$$ where $d$ is the degree. Here we have the values of the $d$ elementary polynomials in the roots $x_1,x_2,\cdots, x_d$ (Vieta's coefficients) which are $e_k=e_k(x_1,x_2,\cdots, x_d)=(-1)^k(k+1)$ and we have five answers to choose the correct:
$$a)\space -(d+1)\hspace1cm b)\space -d\hspace1cm c)\space -(d-1)\hspace1cm d)\space -(d-2)\hspace1cm e)\space -(d-3)$$
The problem being to calculate $$\sum_{n=0}^{d-1}(x_1^n+x_2^n+\cdots+x_d^n)$$
We use Newton's identities to solve the problem for general $d$ degree.
Noting $P_i=x_1^i+x_2^i+\cdots+x_d^i$, Newton's identities give for the five first values
$$\begin{cases}P_1=e_1(x_1,x_2,\cdots, x_d)=e_1\\P_2=e_1^2-2e_2\\P_3=e_1^3-3e_1e_2+3e_3\\P_4=e_1^4-4e_1^2e_2+4e_1e_3+2e_2^2-4e_4\\P_5=e_1^5-5e_2e_1^3+5e_3e_1^2+5e_2^2e_1-5e_4e_1-5e_3e_2+5e_5\end{cases}$$ Calculation gives now a constant value for each $P_i$
$$P_i=-2\text{ for } 1\le i\le 5$$ We assume as a consequence and without proof that $P_i=-2$ for all $i$ which we have also verified for $P_6$ but it is too long to write.
Thus $$\sum_{n=0}^{d-1}(x_1^n+x_2^n+\cdots+x_d^n)= (d)+\underbrace{(-2)+(-2)+\cdots+(-2)}_{(d-1)\space\text{ times}}=d-2(d-1)=-(d-2)$$ Thus in this particular problem the answer is $\boxed{-2013}$ (the option d).
