In the \(n^{\text{th}}\) row of Pascal’s triangle, leave the left \(1\) untouched, multiply the next number along (which is \(\binom{n}{1}=n\)) by \(2\), multiply the next number along (which is \(\binom{n}{2}=\frac{n(n-1}{2}\)) by \(4\), and so on, until you multiply the right-hand \(1\) by \(2^n\). That is, multiply the \(k^{\text{th}}\) number from the left by \(2^k\).
Now what’s the sum of the numbers in the \(n^{\text{th}}\) row?