How to prove $\sum_{i=0}^{n} \binom{n}{i}=2^n$

Question

I need to prove this equality,

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

but I don't know how to start and what even to do

Answer 1

There are at least two ways to prove that, one already shown by the binomial theorem and the other by a counting argument that is:

• consider a set of $n$ elements and count the number of possible subsets of $k\le n$ elements we can choose among them, that is by definition

$$\sum_{k=0}^n \binom nk$$

• now since for each element we have two choice, to include or not include that in a given subset, by the Rule of product the number of all possible subsets with $k\le n$ elements is given by

$$\overbrace{2\cdot 2\cdot \ldots \cdot 2}^{n \,elements}=2^n$$

Answer 2

Use the binomial theorem:

$$\sum_{k=0}^n \binom{n}{k} x^k=(1+x)^n$$

putting $x = 1$,$$ \sum_{k=0}^n \binom nk=2^n$$

Think about the expansion of $(a+b)^n$

$$(a+b)^n = \binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+...+\binom{n}{n-1}ab^{n-1}+\binom{n}{n}b^n $$

Now substitute $a = 1, b = 1$,

$$(1+1)^n = 2^n = \binom{n}{0}1^n+\binom{n}{1}1^{n-1}1+...+\binom{n}{n-1}1.1^{n-1}+\binom{n}{n}1^n = \sum_{k=0}^n \binom{n}{k}$$

Hence,

$$\sum_{k=0}^n \binom{n}{k} = 2^n$$ which is what you wanted.

Answer 3

$$\Sigma_{i=0}^n C_i^n=C_0^n+C_1^n+C_2^n+C_3^n+...+C_n^n$$ you expand using Binomial Theorem:$$(1+x)^n=C_0^n+C_1^n\cdot x+C_2^n\cdot x^2+C_3^n\cdot x^3+...+C_n^n\cdot x^n$$ On substituting $x=1$:
$$(1+1)^n=C_0^n+C_1^n\cdot 1+C_2^n\cdot 1^2+C_3^n\cdot 1^3+...+C_n^n\cdot 1^n$$ Or, $$2^n=C_0^n+C_1^n+C_2^n+C_3^n+...+C_n^n=\Sigma_{i=0}^n C_i^n$$

$$\Sigma_{i=0}^n C_i^n=2^n$$