## Precalculus (10th Edition)

Geometric Sum: $0$
We are given the sequence: $\{(-1)^n\}$ Compute the ratio between two consecutive terms: $\dfrac{a_{k+1}}{a_k}=\dfrac{(-1)^{k+1}}{(-1)^k}=-1$ As the ratio between any consecutive terms is constant, the sequence is geometric. Its elements are: $a_1=(-1)^1=-1$ $r=-1$ We determine the sum of the first 50 terms: $S_n=a_1\cdot\dfrac{1-r^n}{1-r}$ $S_{50}=(-1)\cdot\dfrac{1-\left(-1\right)^{50}}{1-(-1)}=0$