Sum of Binomial Coefficients over Lower Index

Theorem

 * $\displaystyle \sum_{i=0}^n \binom n i = 2^n$

where $\displaystyle \binom n i$ is a binomial coefficient.

Corollary

 * $\displaystyle \sum_{i \in \Z} \binom n i = 2^n$

Proof 1
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{i=0}^n \binom n i = 2^n$

$P(0)$ is true, as this just says $\displaystyle \binom 0 0 = 1$. This holds by definition.

Basis for the Induction
$P(1)$ is true, as this just says $\displaystyle \binom 1 0 + \binom 1 1 = 2$. This also holds by definition.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{i=0}^k \binom {k} {i} = 2^k$

Then we need to show:
 * $\displaystyle \sum_{i=0}^{k+1} \binom {k+1} {i} = 2^{k+1}$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore $\displaystyle \forall n \in \N: \sum_{i=0}^n \binom n i = 2^n$.

Proof 2
Let $S$ be a set with $n$ elements.

From the definition of $r$-combination, $\displaystyle \sum_{i=0}^n \binom n i$ is the total number of subsets of $S$.

Hence $\displaystyle \sum_{i=0}^n \binom n i$ is equal to the cardinality of the power set of $S$.

Hence the result.

Proof 3
From the Binomial Theorem, we have that:


 * $\displaystyle \forall n \in \Z_+: \left({x+y}\right)^n = \sum_{i=0}^n \binom n i x^{n-i}y^i$

Putting $x = y = 1$ we get:

Proof of Corollary
From the definition of the binomial coefficient, when $i < 0$ and $i > n$ we have $\displaystyle \binom n i = 0$.

The result follows directly.