I am running into trouble for proving the inductive conclusion for the proof of the following statement:
For all non-negative integers $n$, $$\sum_{r=0}^n r\binom{n}{r}=\frac{1}{2}n\sum_{r=0}^n \binom{n}{r}$$ Induction on n
Base case $(n=0)$: $$\sum_{r=0}^0 r\binom{0}{r}=0\qquad\frac{1}{2}(0)\sum_{r=0}^0 \binom{0}{r}1=0\qquad 0=0$$ Therefore true when $(n=0)$
Inductive hypothesis: Assume $$\sum_{r=0}^k r\binom{k}{r}=\frac{1}{2}k\sum_{r=0}^k\binom{k}{r}$$ for some $k\in\mathbb{Z}$ where $k\ge 0$
Inductive conclusion: Want to prove $$\sum_{r=0}^{k+1}r\binom{k+1}{r}=\frac{1}{2}(k+1)\sum_{r=0}^{k+1}\binom{k+1}{r}$$ So $$\sum_{r=0}^{k+1} r\binom{k+1}{r}=\sum_{r=0}^{k} r\binom{k+1}{r}+(k+1)\binom{k+1}{k+1}=\sum_{r=0}^{k} r\binom{k+1}{r}+(k+1)$$ So yeah, this is where I'm stuck. Any help?