If a and b are real numbers and n is an integer then the expression $(a+b)^n$ can be expanded as:

$ (a+b)^n = a^n+ {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^ + \ldots +{n \choose n-1}ab^{n-1}+b^n$

$ (a+b)^n = \sum_{i=1}^n {n \choose x}a^xb^{n-x}$

This can be seen by looking at $(a+b)^ n$ as a product of n parentheses and multiply these by picking one item (a or b) from each. If we picked $a$ from x parentheses and $b$ from $(n-x)$, then the product is $a^x b^{n-x}$. We can choose the $x$ $a$'s in a total of $\binom{n}{x}$ ways so the coefficient of $a^xb^{n-x}$ is $\binom{n}{x}$.

\begin{xmpl}
Since
$$(a+b)^n = \sum_{x=0}^n {n \choose x} a^xb^{n-x},$$
it follows that
$$2^n = (1+1)^n = \sum_{x=0}^n {n \choose x}$$
i.e.
$$2^n = {n \choose 0} + {n \choose 1} + {n \choose 2}\ldots+ {n \choose n}$$

\end{xmpl}