<s>[INST] Stochastics with induction 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?
 [/INST] 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.
</s>