Working with Geometric Series - AP Calculus BC

Compute $\sum_{n=2}^{\infty} 6\left(\tfrac{1}{2}\right)^n$ by treating it as a geometric series.
Answer: $\tfrac{6\left(\tfrac{1}{2}\right)^2}{1-\tfrac{1}{2}}$ This problem involves a geometric series starting at n=2, requiring careful handling of the initial index. The series ∑(n=2 to ∞) 6(1/2)ⁿ equals 6(1/2)² + 6(1/2)³ + ... = 6(1/2)²[1 + 1/2 + (1/2)² + ...]. Factoring out 6(1/2)², we get a standard geometric series with first term a = 1 and ratio r = 1/2. Since |r| = 1/2 < 1, this inner series sums to 1/(1-1/2) = 2, giving us a total of 6(1/2)² · 2 = 6(1/4) · 2 = 3/2. Alternatively, we can compute directly as 6(1/2)²/(1-1/2) = (6/4)/(1/2) = 3/2. Choice A incorrectly treats it as starting from n=0, while choice D wrongly claims divergence. For series starting at n=k with k>0, factor out the common terms to reduce to a standard geometric series.
Evaluate $\sum_{n=1}^{\infty} 6\left(-\dfrac{1}{3}\right)^{n-1}$ , or state that it diverges.
Answer: $\dfrac{9}{2}$ This problem asks us to evaluate the geometric series Σ(n=1 to ∞) 6 $(-1/3)^{n-1}$ . Since the exponent is n-1 and n starts at 1, the first term is a = 6 $(-1/3)^0$ = 6, and the common ratio is r = -1/3. Since |r| = |-1/3| = 1/3 < 1, the series converges. Using the sum formula S = a/(1-r), we get S = 6/(1-(-1/3)) = 6/(1+1/3) = 6/(4/3) = 18/4 = 9/2. Choice E shows the correct formula structure 6/(1+1/3) but written differently. For geometric series with negative ratios, remember that the sum formula becomes a/(1-r) where subtracting a negative r adds to 1.
A light's intensity is multiplied by $-0.6$ each reflection, starting at 10 units; what is the infinite sum of intensities?
Answer: $6.25$ This problem involves finding the sum of an infinite geometric series of light intensities. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 10 and r = -0.6, so |r| = 0.6 < 1, ensuring convergence. The sum is S = a / (1 - r) = 10 / (1 - (-0.6)) = 10 / 1.6 = 6.25. One tempting distractor is -6.25, which might result from using 1 + r instead of 1 - r in the denominator. When working with geometric series in iterative processes like reflections, verify the sign of the ratio to apply the sum formula correctly.
Determine the value of $\sum_{n=1}^{\infty} \tfrac{4}{3^n}$ or state that it diverges.
Answer: $2$ This problem involves finding the sum of an infinite geometric series. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the series is $4 \times \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n$ , with $r = \frac{1}{3} < 1$ , ensuring convergence. The sum is $4 \times \frac{\frac{1}{3}}{1 - \frac{1}{3}} = 4 \times \frac{\frac{1}{3}}{\frac{2}{3}} = 4 \times \frac{1}{2} = 2$ . One tempting distractor is 4, which could come from mistakenly including the n=0 term or misapplying the formula. When summing geometric series starting from n=1, remember to use the formula for the tail sum appropriately.
An artist adds $3$ mL of dye, then $-\tfrac{1}{2}$ times the previous amount repeatedly; what is the infinite sum?
Answer: $2$ This problem requires working with geometric series to find the total dye added by the artist. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the amounts start with 3 mL and continue with r = -1/2, and since |-1/2| < 1, the series converges. The sum is S = 3 / (1 - (-1/2)) = 3 / (3/2) = 2. One tempting distractor is 6, which might come from incorrectly using r = 1/2 without the negative sign, leading to 3 / (1/2) = 6. A transferable strategy for geometric series is to identify the first term and common ratio, confirm |r| < 1 for convergence, and apply the sum formula, paying attention to signs for alternating series.
A medication dose starts at $80$ mg and decreases by $20\%$ each hour. What total amount is administered over all hours?
Answer: $\tfrac{80}{1-\tfrac{4}{5}}$ This medication problem creates a geometric series from a real-world scenario. Starting with 80 mg and decreasing by 20% each hour means each dose is 80% = 0.8 = 4/5 of the previous dose. The total amount is 80 + 80(4/5) + 80(4/5)² + ... = 80[1 + 4/5 + (4/5)² + ...], which is a geometric series with first term a = 80 and common ratio r = 4/5. Since |r| = 4/5 < 1, the series converges to a/(1-r) = 80/(1-4/5) = 80/(1/5) = 400. Choice B incorrectly uses 1/5 as the ratio instead of 4/5, while choice E uses the wrong sign. When translating percentage decreases to geometric series, remember that "decreases by 20%" means "multiplies by 0.8" for the common ratio.
A ball travels $12$ m, then each bounce travels $\tfrac{3}{4}$ of the previous distance. What is the total distance?
Answer: $48$ This problem requires finding the sum of a geometric series representing the ball's total travel distance. The ball travels 12 m initially, then 12(3/4) m, then 12(3/4)² m, and so on, giving us the series 12 + 12(3/4) + 12(3/4)² + ... = 12[1 + 3/4 + (3/4)² + ...]. This is a geometric series with first term a = 12 and common ratio r = 3/4. Since |r| = 3/4 < 1, the series converges to a/(1-r) = 12/(1-3/4) = 12/(1/4) = 48. Choice D incorrectly uses only the formula without the initial term factored out, while choice E uses the wrong sign in the denominator. When working with geometric series in applications, always identify the first term and common ratio, then apply the formula a/(1-r) for |r| < 1.
Determine the sum of the convergent series $\sum_{n=0}^{\infty} \tfrac{3}{8}\left(\tfrac{2}{3}\right)^n$ .
Answer: $\tfrac{3}{8}\cdot\tfrac{1}{1-\tfrac{2}{3}}$ This problem asks for the sum of a straightforward geometric series. The series ∑(n=0 to ∞) (3/8)(2/3)ⁿ has first term a = 3/8 (when n=0) and common ratio r = 2/3. Since |r| = 2/3 < 1, the series converges to a/(1-r) = (3/8)/(1-2/3) = (3/8)/(1/3) = (3/8)·3 = 9/8. This can also be written as (3/8)·(1/(1-2/3)) as shown in choice A. Choice B incorrectly uses 3/2 as the ratio (the reciprocal), while choice E unnecessarily includes an extra factor of 2/3. For geometric series of the form c·rⁿ starting at n=0, the sum is simply c/(1-r) when |r| < 1.
Evaluate $\sum_{n=0}^{\infty} 5\left(-\tfrac{2}{3}\right)^n$ by determining convergence and its sum.
Answer: $3$ This problem asks us to evaluate a geometric series with alternating signs. The series ∑(n=0 to ∞) 5(-2/3)ⁿ has first term a = 5(-2/3)⁰ = 5 and common ratio r = -2/3. Since |r| = |-2/3| = 2/3 < 1, the series converges. Using the geometric series formula, the sum is a/(1-r) = 5/(1-(-2/3)) = 5/(1+2/3) = 5/(5/3) = 5·(3/5) = 3. Choice E shows the correct formula setup but doesn't simplify, while choice D incorrectly uses only the first term. For geometric series with negative ratios, remember that convergence depends on |r| < 1, and the sum formula uses 1-r with the actual signed value of r.
A light's intensity is $12$ units, then multiplies by $1.05$ each second; does the infinite total intensity converge?
Answer: Diverges because $|r|>1$ This problem involves working with geometric series to determine if the total light intensity converges. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the intensities form a series with first term 12 and r = 1.05, but |1.05| > 1, so the series diverges. The sum formula S = a / (1 - r) only applies when |r| < 1, which is not the case here. One tempting distractor is converging to 12/(1-1.05), which fails because the condition |r| < 1 is not met, making the formula invalid. A transferable strategy for geometric series is to identify the first term and common ratio, check the convergence condition |r| < 1, and only then apply the sum formula if appropriate.
A pattern uses areas $9, -18, 36, -72, \dots$ ; does the infinite series converge or diverge?
Answer: Diverges because $|r|>1$ This problem involves working with geometric series to determine if the sum of areas converges. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the areas are 9, -18, 36, -72, ..., with r = -2, but |-2| > 1, so the series diverges. The sum formula S = a / (1 - r) only applies when |r| < 1, which is not satisfied here. One tempting distractor is converging to 9/(1-(-2)), which fails because |r| > 1 violates the convergence condition, making the sum infinite. A transferable strategy for geometric series is to identify the first term and common ratio, check the convergence condition |r| < 1 first, and only apply the sum formula if the series converges.
A savings plan deposits $200$ dollars now, then each month deposits $90\%$ of the previous deposit. What is the total deposited?
Answer: $\tfrac{200}{1-0.9}$ This savings problem creates a geometric series from successive deposits. Starting with $200 and depositing 90% = 0.9 of the previous amount each month gives the series 200 + 200(0.9) + 200(0.9)² + ... = 200[1 + 0.9 + 0.9² + ...]. This is a geometric series with first term a = 200 and common ratio r = 0.9. Since |r| = 0.9 < 1, the series converges to a/(1-r) = 200/(1-0.9) = 200/0.1 = 2000. Choice C incorrectly uses 0.1 directly instead of recognizing that 1-0.9 = 0.1, while choice D wrongly claims divergence for a convergent series. When working with percentage-based geometric series, convert percentages to decimals and apply the formula a/(1-r) when |r| < 1.
Find the sum of $1-\tfrac{1}{3}+\tfrac{1}{9}-\tfrac{1}{27}+\cdots$ if it converges.
Answer: $\tfrac{1}{1-\left(-\tfrac{1}{3}\right)}$ This alternating series problem requires recognizing the pattern as a geometric series. The series 1 - 1/3 + 1/9 - 1/27 + ... has terms that alternate in sign with pattern 1, -1/3, 1/9, -1/27, which equals 1, (-1/3)¹, (-1/3)², (-1/3)³, and so on. This is a geometric series with first term a = 1 and common ratio r = -1/3. Since |r| = |-1/3| = 1/3 < 1, the series converges to a/(1-r) = 1/(1-(-1/3)) = 1/(1+1/3) = 1/(4/3) = 3/4. Choice A incorrectly ignores the negative sign in the ratio, while choice C uses the wrong formula structure. For alternating geometric series, identify the signed common ratio and apply the standard formula with that signed value.
Determine whether $\sum_{n=0}^{\infty} \left(\tfrac{5}{2}\right)^n$ converges, and select the correct conclusion.
Answer: Diverges because $\left|\tfrac{5}{2}\right|>1$ This problem tests understanding of the convergence condition for geometric series. The series ∑(n=0 to ∞) (5/2)ⁿ has first term a = 1 and common ratio r = 5/2. For a geometric series to converge, we need |r| < 1. Here, |r| = |5/2| = 5/2 = 2.5 > 1, so the series diverges. The terms (5/2)ⁿ grow without bound as n increases, preventing the series from having a finite sum. Choice A incorrectly attempts to apply the sum formula despite divergence, while choice D gives the wrong reason for divergence. Remember that geometric series converge if and only if |r| < 1; when |r| ≥ 1, the series diverges regardless of the sign of r.
Evaluate $\sum_{n=0}^{\infty} 5\left(\tfrac{2}{3}\right)^n$ or state that it diverges.
Answer: $15$ This problem involves finding the sum of an infinite geometric series. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 5 and r = 2/3, so |r| = 2/3 < 1, ensuring convergence. The sum is given by S = a / (1 - r) = 5 / (1 - 2/3) = 5 / (1/3) = 15. One tempting distractor is 15/2, which might result from mistakenly using the formula for the sum starting from n=1 instead of n=0, excluding the initial term. When working with geometric series, always identify the first term and common ratio accurately to apply the sum formula correctly.
A sequence of lengths is $12+6+3+\cdots$ centimeters. What is the sum of the infinite series?
Answer: $24$ This problem involves finding the sum of an infinite geometric series of lengths. The series is 12 + 6 + 3 + ..., where the first term a = 12 and the common ratio r = 6/12 = 1/2 = 0.5. Since |r| = 0.5 < 1, the series converges. Applying the sum formula S = a/(1-r), we get S = 12/(1-0.5) = 12/0.5 = 24. Choice E (12/(1-2)) would give a negative denominator, which makes no sense for a convergent series. Choice A (18) might result from arithmetic errors. The key to geometric series problems is methodically finding r, checking convergence, and carefully applying the sum formula.
A ball travels $24$ ft, then each bounce travels $\tfrac{2}{3}$ as far; what total distance does it travel?
Answer: $72$ This problem requires working with geometric series to find the total distance traveled by a bouncing ball. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the series is formed by the initial travel of 24 ft followed by subsequent travels scaled by r = 2/3 each time. Since |2/3| < 1, the series converges, and the sum is given by the formula S = a / (1 - r), where a = 24, yielding S = 24 / (1/3) = 72. One tempting distractor is 48, which might arise from incorrectly doubling only part of the series or misidentifying the first term. A transferable strategy for geometric series is to identify the first term and common ratio, confirm |r| < 1 for convergence, and apply the sum formula while considering the physical context.
Find the sum of $\sum_{n=0}^{\infty} \left(-\tfrac{3}{4}\right)^n$ or state that it diverges.
Answer: $\tfrac{4}{7}$ This problem involves finding the sum of an infinite geometric series. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 1 and r = -3/4, so |r| = 3/4 < 1, ensuring convergence. The sum is S = a / (1 - r) = 1 / (1 - (-3/4)) = 1 / (7/4) = 4/7. One tempting distractor is 7/4, which could come from inverting the fraction incorrectly or confusing the formula. When evaluating alternating geometric series, double-check the denominator to ensure it's 1 minus the ratio.
A computer renders $80$ frames, then each pass renders $\tfrac{3}{8}$ as many; what is the total frames rendered?
Answer: $128$ This problem involves working with geometric series to find the total frames rendered by a computer. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the frames start with 80 and continue with $r = \tfrac{3}{8}$ , and since $|\tfrac{3}{8}| < 1$ , the series converges. The sum is $S = 80 / (1 - \tfrac{3}{8}) = 80 / (\tfrac{5}{8}) = 128$ . One tempting distractor is $80/(1-\tfrac{8}{3})$ , which fails because it reverses the ratio fraction, leading to a negative and incorrect sum. A transferable strategy for geometric series is to identify the first term and common ratio, confirm $|r| < 1$ for convergence, and apply the sum formula accurately.
Compute $\sum_{n=3}^{\infty} 2\left(\tfrac{1}{4}\right)^n$ or state that it diverges.
Answer: $\tfrac{1}{24}$ This problem involves finding the sum of an infinite geometric series starting from a specific term. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, r = 1/4, so |r| = 1/4 < 1, ensuring convergence, and the sum from n=3 to infinity is 2* $(1/4)^3$ / (1 - 1/4) = (2/64) / (3/4) = (1/32) * (4/3) = 1/24. This can be derived by factoring out the initial terms to form a standard geometric sum. One tempting distractor is 1/32, which is just the first term without summing the rest of the series. When summing geometric series from a non-zero index, express it as the full sum minus the finite initial partial sum or factor appropriately.

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