prove that for all $n \in \mathbb{N}$:
$\sum_{r=0}^n \binom{n}{r}(-1)^{r} = 0$.
The base step is easy, i only get lots of problems when i try to mess with the sum boundries....
so far i've tried:
$\sum_{r=0}^{n+1} \binom{n+1}{r}(-1)^{r} = \sum_{r=0}^{n+1}(\binom{n}{r}+\binom{n}{r-1})(-1)^{r} = \sum_{r=0}^{n+1}\binom{n}{r}(-1)^{r}+\sum_{r=0}^{n+1}\binom{n}{r-1}(-1)^{r}$
Don't know how to proceed... Please help?
Perhaps this way: $$ \begin{split} \sum_{r=0}^{n+1} \binom{n+1}{r} (-1)^r &= \binom{n+1}{0} + \binom{n+1}{n+1} (-1)^{n+1} + \sum_{r=1}^n \binom{n+1}{r} (-1)^r \\ &= 1 + (-1)^{n+1} + \sum_{r=0}^{n-1} \binom{n+1}{r+1} (-1)^{r+1} \\ &= 1 + (-1)^{n+1} - \sum_{r=0}^{n-1} \binom{n+1}{r+1} (-1)^r \\ &= 1 + (-1)^{n+1} - \sum_{r=0}^{n-1} \binom{n}{r+1} (-1)^r - \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\ &= 1 + (-1)^{n+1} + \sum_{r=0}^{n-1} \binom{n}{r+1} (-1)^{r+1} - \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\ &= 1 + (-1)^{n+1} + \sum_{r=1}^n \binom{n}{r} (-1)^r - \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\ & \text{cancel terms in the sum indexed $1...n-1$} \\ &= 1 + (-1)^{n+1} + \binom{n}{n}(-1)^n- \binom{n}{0} \\ &= (-1)^{n+1} + (-1)^n \\ &= 0 \end{split} $$ as desired.