Sum of Binomial Coefficients over Lower Index

Theorem

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

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

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


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

Basis for the Induction

 * $$P(1)$$ is true, as this just says $$\binom 1 0 + \binom 1 1 = 2$$. 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:

$$\sum_{i=0}^k \binom {k} {i} = 2^k$$.

Then we need to show:

$$\sum_{i=0}^{k+1} \binom {k+1} {i} = 2^{k+1}$$.

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

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

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

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

Hence the result.

Proof 3
From the Binomial Theorem, we have that:


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

Putting $$x = y = 1$$ we get:

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