How would you find the sum \(\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + ... + \binom{n}{n}\) from Pascal’s triangle? Do so for \(n\) = \(2\), \(3\), and \(4\). Guess at the answer in general.

For \(n = 2\) it’s \(4\), for \(n = 3\) it’s \(8\), for \(n = 4\) it’s \(16\) so it certainly looks like the formula is \(2^n\)