Let $u_r = {2N-r \choose N } 2^{-2N+r}$ for $r = 0, 1, 2 , ... , N$. It has been shown that $\sum_{r=0}^N u_r =1$, so $u_r$ is the probability distribution. I need to show that the expectation of the distribution $\{u_r\}$ is $(2N+1)u_0-1$. The hint suggests to use the relation $$(N-r)u_r = \frac12(2N+1)u_{r+1} - \frac12(r+1)u_{r+1}.$$
Let the expectation of the distribution be $\mu$. Then,
$$\mu = \sum_{r=0}^N r u_r= - \left(\sum_{r=0}^N (N-r) u_r - \sum_{r=0}^NNu_r\right).$$ However, applying the hint to the right hand side doesn't seem to be helpful. Could you give some hint?
The second term is just $N$ (due to the normalization). If you use the hint in the first term, you get another sum involving only $u_{r+1}$ and not $r$, which you can evaluate using the normalization (taking care with the limits), and a sum over $(r+1)u_{r+1}$, which you can relabel as $ru_r$ (again taking care with the limits), so this yields a multiple of the desired expectation. Then you can solve the resulting linear equation for the expectation.