Question 837300
Break up the sum as follows:


*[tex \large \sum_{k=1}^{n} 2^{k-1}k = \sum_{k=1}^n 2^{k-1} + \sum_{k=2}^{n} 2^{k-1} + \ldots + \sum_{k = n}^n 2^{k-1}]


Each of the summations on the RHS is a geometric series, which can be evaluated with some algebra.