What is wrong with this proof about a sum of binomial coefficients?

Question

I want to show that $\sum_{k=0}^n\binom{n+1}{k}=2^{n+1}-1$

Proof:

$$\sum_{k=0}^n\binom{n+1}{k} = \sum_{k=-1}^{n-1}\binom{n+1}{k+1} \;\; q:=k+1, \; r:=n+1$$ thus the sum equals $$\sum_{q=0}^{r-2}\binom{r}{q} = \sum_{q=0}^{r}\binom{r}{q} -\binom{r}{r-1} -\binom{r}{r} = 2^{r}-r-1 = 2^{n+1}-n-2$$ so clearly I did something wrong.

I tried to double-check everything but all steps seem OK to me.

Where did I go wrong?

Answer 1

You made a mistake in your substitutions. If you let $q=k+1$, your lower limit is indeed $q=-1+1=0$. But your upper limit should be $q=(n-1)+1=n=r-1$ (instead of $r-2$). So you have

\begin{align}\sum_{q=0}^{r-1}\binom{r}{q} = \sum_{q=0}^{r}\binom{r}{q}-\binom{r}{r}=2^r-1=2^{n+1}-1. \end{align}