How do I even pronounce this word?

Let us solve the general problem, where each term in the sequence is the sum of previous $n$ terms. Then from the theory of linear recurrence it follows that $R$ is a solution of the following equation $x^n=x^{n-1}+x^{n-2} + \ldots + x +1$ Hence it is clear that $f \equiv x^n-x^{n-1}-x^{n-2} -\ldots -x -1$ and $\text{deg}(f)=n$ . Thus our objective is to evaluate $\sum_{\alpha}\alpha^n$ Where $\alpha$ is a root of $f$ . Now multiplying both sides of the above equation by $x-1$ and introducing an extraneous root 1, the above equation simplifies to $x^{n+1}-2x^n+1=0\hspace{20pt}(1)$ Which immediately yields, $x^n=\frac{1}{2-x} \hspace{20pt} (2)$ Substituting $z=\frac{1}{2-x}$ and hence $x=2 -\frac{1}{z}$ in Equation (1), we have $(2-\frac{1}{z})^{n+1}-2(2-\frac{1}{z})^n + 1=0$ Simplifying (using binomial theorem), $z^{n+1}-2^nz^{n}+ \ldots =0$ Hence sum of the roots of the above equation is equal to $2^n$ . By Equation (2), this sum is precisely $\sum_{\alpha}\alpha^n$ plus the extraneous root $1$ that we introduced earlier. Hence the required answer is $2^n-1$ which is 2047 for $n=11$ in the problem. $\blacksquare$

Alright, so we know $f(x) = x^{11}-x^{10}-\ldots-x-1$ . Newton's sums states that

$\begin{array}{c}\\ P_{1}-1 & = 0 \\ P_{2}-P_{1}-2 & = 0 \\ P_{3}-P_{2}-P_{1}-3 & = 0 \\ \ldots \\ P_{k}-P_{k-1}-\ldots-P_{1}-k & = 0 \end{array}$

for all $k \leq 11$ , where $\displaystyle P_{k} = \sum r_{i}^{k}$ .

After computing up to $P_{4}$ or so, we notice that they looks like $2^{k}-1$ . To confirm our suspicions, we will proceed with induction.

Clearly, for the base case $k = 1$ , $P_{k} = 2^{k}-1$ holds true. Now, we go about inducting:

$P_{k}-P_{k-1}-\ldots-P_{1}-k = 2^{k}-1-(2^{k-1}-1-\ldots+2^{1}-1)-k$

$= 2^{k}-1-(2^{k}-2-k+1)-k = 0$

And thus, we get that

$\sum_{f(r_{i}) = 0} r_{i}^{\deg(f)} = P_{11} = 2^{11}-1 = \boxed{2047}$