Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Review Exercises - Page 839: 16

$682$

We have to determine the sum: $\sum_{k=1}^{10} (-2)^k$ Consider the geometric sequence: $a_1=(-2)^1=-2$ $r=-2$ Compute the sum of the first 10 terms: $S_n=a_1\dfrac{1-r^n}{1-r}$ $n=10$ $\sum_{k=1}^{10} (-2)^k=S_7=(-2)\cdot\dfrac{1-(-2)^{10}}{1-(-2)}$ $=-2\cdot \dfrac{1-1024}{3}=-2\cdot\dfrac{-1023}{3}=2(341)=682$