Prove the following combination?

Question

I need a quick proof of the following combination

$$\binom{n+1}1+\binom{n+1}2+\binom{n+1}3+\dots++\binom{n+1}{n+1}=2^{n+1}-1$$

Answer 1

Daniel Fischer has already given you a very nice answer. However, it might be nice to see an alternative more combinatorial approach.

Imagine you are trying to compute the ways you can form a subset from $n+1$ objects, without that subset being empty. It should be the ways you can choose $1$ object from among those $n+1$ objects, plus the ways you can choose $2$ objects from among those $n+1$ objects, and so on up until you are choosing all $n+1$ objects to form your subset.

We can compute this sum (that is, the number of subsets we can form) by imagining a subset as a bit string of length $n+1$, where each entry is $0$ if the object is not included in the subset, and each entry is a $1$ if that object IS included in the subset. There are clearly $2^{n+1}$ such bit strings, but we must subtract $1$, as we cannot include the bit string where all entries are $0$, i.e. the empty set.

Hence, we arrive exactly at your identity.