Proving a binomial series to be $2^n$

Question

I tried to prove $$\sum\limits_{r=0}^n\binom{n}{r}=\sum\limits_{r=0}^n \frac{n!}{r!(n-r)!} =2^n.$$

I used induction method:

Assuming $$\sum\limits_{r=0}^n \frac{n!}{r!(n-r)!} =2^{n}.$$

Proving $$\sum\limits_{r=0}^{n+1} \frac{{(n+1)}!}{r!(n+1-r)!} =2^{n+1}.$$

I got $$\sum\limits_{r=0}^{n+1} \frac{{(n+1)}!}{r!(n+1-r)!} =\sum\limits_{r=0}^{n+1} \Bigl(\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}\Bigr),$$ and an invalid term appeared: $(-1)!$.

How does one prove this identity, and what's wrong with my method?

As an aside, rather than writing things out with factorials, you can use binomial coefficients instead. Now, note that $\binom{n}{r} = 0$ for all cases where $r<0$ or $r>n$. — JMoravitz, Dec 14 '18 at 04:27

score 0 · Answer 1 · answered Dec 14 '18 at 04:23

The identity you are using in the last equality has some hypothesis: it holds when $r > 0$.

There are many approaches to prove this fact. I particulary like a combinatorial proof via the following counting argument: if $A_k$ is the set of subsets of $\{1, \dots, n\}$ of size $k$, then by (one possible) definition, $|A_k| = {n \choose k}$. Moreover, we have that the subsets of $\{1, \dots, n\}$, that is the elements of $\mathcal{P}(\{1,\dots,n\})$, are the disjoint union of $A_0, \dots, A_n$.

Thus, $\mathcal{P}(\{1,\dots,n\}) = \sqcup_{0 \leq i \leq n}A_i$ and

$$ 2^n = |\mathcal{P}(\{1,\dots,n\})| = |\sqcup_{0 \leq i \leq n}A_i| = \sum_{i=0}^n|A_i| = \sum_{i = 0}^n{n \choose i}. $$

Proving a binomial series to be $2^n$

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