## College Algebra (10th Edition)

$1-2^{n}$
There is a common ratio between terms, $r=2.$ The terms of the sum form a geometric sequence, $a_{1}=-1,\ r=2.$ We can write the sum as $S=\displaystyle \sum_{k=1}^{n}(-1)\cdot 2^{k-1}$ Apply THEOREM: Sum of the First $n$ Terms of a Geometric Sequence $S_{n}=\displaystyle \sum_{k=1}^{n}a_{1}r^{k-1}=a_{1}\cdot\frac{1-r^{n}}{1-r},\quad r\neq 0,1$ $=-1\displaystyle \left(\frac{1-2^{n}}{1-2}\right)$ $=-1\displaystyle \left(\frac{1-2^{n}}{-1}\right)$ $=1-2^{n}$