1. **Stating the problem:** We want to find the sum $S_5 = b_1 + b_2 + b_3 + b_4 + b_5$ where the terms are given as $b_k = (-3)^{k-1}$ for $k=1$ to $5$. 2. **Formula used:** This is a geometric series with first term $a = 1$ (since $(-3)^0 = 1$) and common ratio $r = -3$. The sum of the first $n$ terms of a geometric series is given by: $$S_n = a \frac{1-r^n}{1-r}$$ 3. **Applying the formula:** Here, $n=5$, $a=1$, and $r=-3$. $$S_5 = 1 \times \frac{1 - (-3)^5}{1 - (-3)} = \frac{1 - (-243)}{1 + 3} = \frac{1 + 243}{4}$$ 4. **Simplifying:** $$S_5 = \frac{244}{4}$$ 5. **Final answer:** $$S_5 = 61$$ So, the sum of the series is 61.