Show that $$\Sigma_{r=0}^{n}(2r+1)P_r(x)=P_{n+1}'(x)+P_n'(x).$$
I've started with the relation $$P_l'(x)=\Sigma_{r=0}^{\frac{1}{2}(l-1)}(2l-4r-1)P_{l-2r-1}(x).$$ I tried adding the sums for $P_n'(x)$ and $P_{n+1}'(x)$ to show that this is equal to the first expression, but that didn't get me anywhere. Any ideas?
Even simpler: Bonnet's recursion formula gives
$$ (2n+1) P_n(x) = P_{n+1}'(x)-P_{n-1}'(x) $$ hence the given sum is a telescopic sum.