$$ [-1,-4, ... ,-31; \frac{x^{31}}{x-1}]$$
Note: $$ [x_0,x_1,...,x_n;f] = \sum_{i=0}^{n} \frac{f(x_i)}{l'(x_i)} $$
where $ l(x_k) = (x-x_0)...(x-x_k) $
I know that : $$ [x_0, x_1, ... , x_n; \frac{f(x)}{x-a}] = [x_0, x_1, ... , x_n,a; f(x)] - \frac{f(a)}{(a-x_0)...(a-x_n)} $$ In this case : $ f(x) = x^{31} , a=1 , x_i = -(3i+1) $ $$ [-1,-4, ... ,-31; \frac{x^{31}}{x-1}] = [-1,-4, ... ,-31,1; x^{31}] - \frac{1}{\prod_{k=0}^{10} (3k+2)} $$
My problem is evaluating : $$[-1,-4, ... ,-31,1; x^{31}]$$ because the grade of the polynomial is $31$ and I have only $11$ nodes .