Sum of Geometric Series - College Algebra

1. A certain bacteria culture doubles every 3 hours. If the culture starts with 200 bacteria, the population after $3n$ hours forms a geometric sequence: 200, 400, 800, 1600, .... What is the sum of the first 6 terms of this sequence?
   Answer: $S_6 = 12600$ The geometric sequence has first term $a = 200$ and common ratio $r = 2$ (since the population doubles). Using the formula $S_n = a \cdot \frac{r^n - 1}{r - 1}$: $S_6 = 200 \cdot \frac{2^6 - 1}{2 - 1} = 200 \cdot \frac{64 - 1}{1} = 200 \cdot 63 = 12600$. We can verify: the first 6 terms are 200, 400, 800, 1600, 3200, 6400, and $200 + 400 + 800 + 1600 + 3200 + 6400 = 12600$.
2. Consider the series $\sum_{k=2}^{7} 3 \cdot 2^{k-2}$. This can be rewritten as a geometric series. What is its sum?
   Answer: $S = 189$ Let's rewrite this series by substituting $j = k - 2$, so when $k = 2$, $j = 0$ and when $k = 7$, $j = 5$. The series becomes $\sum_{j=0}^{5} 3 \cdot 2^j$. This is a geometric series with first term $a = 3 \cdot 2^0 = 3$, common ratio $r = 2$, and 6 terms (from $j = 0$ to $j = 5$). Using $S_n = a \cdot \frac{r^n - 1}{r - 1}$: $S_6 = 3 \cdot \frac{2^6 - 1}{2 - 1} = 3 \cdot \frac{64 - 1}{1} = 3 \cdot 63 = 189$. Alternatively, we can compute directly: the terms are $3 \cdot 2^0 = 3$, $3 \cdot 2^1 = 6$, $3 \cdot 2^2 = 12$, $3 \cdot 2^3 = 24$, $3 \cdot 2^4 = 48$, $3 \cdot 2^5 = 96$. Sum: $3 + 6 + 12 + 24 + 48 + 96 = 189$. Choice B (192) might result from including an extra term or a calculation error. Choice C (195) and D (198) are further off and likely represent other common errors.
3. For what value of $x$ does the infinite geometric series $\sum_{n=0}^{\infty} (2x-1) \cdot \left(\frac{x+1}{3}\right)^n$ converge, and what is its sum when it converges?
   Answer: Converges for $-4 < x < 2$; sum is $\frac{3(2x-1)}{2-x}$ This is an infinite geometric series with first term $a = 2x-1$ and common ratio $r = \frac{x+1}{3}$. For convergence, we need $|r| < 1$, which means $\left|\frac{x+1}{3}\right| < 1$. This gives $|x+1| < 3$, so $-3 < x+1 < 3$, which means $-4 < x < 2$. When the series converges, its sum is $S = \frac{a}{1-r} = \frac{2x-1}{1-\frac{x+1}{3}} = \frac{2x-1}{\frac{3-(x+1)}{3}} = \frac{2x-1}{\frac{2-x}{3}} = \frac{3(2x-1)}{2-x}$.
4. An infinite geometric series has first term $a = 18$ and sum $S = 24$. What is the common ratio $r$?
   Answer: $r = \frac{1}{4}$ For an infinite geometric series with $|r| < 1$, the sum is $S = \frac{a}{1-r}$. Substituting the given values: $24 = \frac{18}{1-r}$. Solving for $r$: $24(1-r) = 18$, so $24 - 24r = 18$, which gives $24r = 6$, and therefore $r = \frac{1}{4}$. We can verify: $|\frac{1}{4}| < 1$ (convergence condition satisfied) and $S = \frac{18}{1-\frac{1}{4}} = \frac{18}{\frac{3}{4}} = 24$ ✓. Choice B gives $r = \frac{1}{3}$: $S = \frac{18}{1-\frac{1}{3}} = \frac{18}{\frac{2}{3}} = 27 \neq 24$. Choice C gives $r = \frac{3}{4}$: $S = \frac{18}{1-\frac{3}{4}} = \frac{18}{\frac{1}{4}} = 72 \neq 24$. Choice D gives $r = \frac{2}{3}$: $S = \frac{18}{1-\frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54 \neq 24$.

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