This sum can be written as $$x=\sum_{k=1}^{s}\sum_{p=1}^kk(s+1)^{p-1}=-\frac{(1+s)(-2+s^3-2(1+s)^s(-1+s^2))}{2s^3}$$ Now you have $2$ cases:
When $s=2k,k\in\mathbb{N}$ after a computation you will get $$x\equiv1\mod{s-1}$$ When $s=2k-1$ you will get $$x\equiv\frac{s+1}{2}\mod s-1$$