Calculus 2 Quiz: Geometric Series
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Geometric SeriesQuestion 1 of 20

Two geometric series are given by SA=∑n=0∞xnS_A = \sum_{n=0}^{\infty} x^n and SB=∑n=0∞ynS_B = \sum_{n=0}^{\infty} y^n. If SA=2S_A=2 and SB=3S_B=3, what is the sum of the series ∑n=0∞(xy)n\sum_{n=0}^{\infty} (xy)^n?

65\frac{6}{5}
32\frac{3}{2}
5
6
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Calculus 2 Quiz

Calculus 2 Quiz: Geometric Series

Practice Geometric Series in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Geometric Series, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two geometric series are given by SA=∑n=0∞xnS_A = \sum_{n=0}^{\infty} x^n and SB=∑n=0∞ynS_B = \sum_{n=0}^{\infty} y^n. If SA=2S_A=2 and SB=3S_B=3, what is the sum of the series ∑n=0∞(xy)n\sum_{n=0}^{\infty} (xy)^n?

  1. 65\frac{6}{5}
  2. 32\frac{3}{2} (correct answer)
  3. 5
  4. 6
Explanation: First, find the values of xx and yy. For series A, SA=11−x=2S_A = \frac{1}{1-x} = 2. This gives 1=2−2x1 = 2-2x, so 2x=12x=1 and x=1/2x=1/2. For series B, SB=11−y=3S_B = \frac{1}{1-y} = 3. This gives 1=3−3y1 = 3-3y, so 3y=23y=2 and y=2/3y=2/3. The third series is ∑n=0∞(xy)n\sum_{n=0}^{\infty} (xy)^n. This is a geometric series with ratio r=xyr = xy. The product xy=(1/2)(2/3)=1/3xy = (1/2)(2/3) = 1/3. Since ∣xy∣=1/3<1|xy| = 1/3 < 1, the series converges. Its sum is 11−xy=11−1/3=12/3=32\frac{1}{1-xy} = \frac{1}{1-1/3} = \frac{1}{2/3} = \frac{3}{2}.

Question 2

The series ∑n=1∞2(k+1)n\sum_{n=1}^{\infty} \frac{2}{(k+1)^n} converges to 1/31/3. What is the value of kk?

  1. 3
  2. 5
  3. 6 (correct answer)
  4. 7
Explanation: The given series is 2∑n=1∞(1k+1)n2 \sum_{n=1}^{\infty} (\frac{1}{k+1})^n. This is a geometric series with common ratio r=1k+1r = \frac{1}{k+1}. Since the series starts at n=1n=1, the first term is a=1k+1a = \frac{1}{k+1}. The sum of this part of the series is S=a1−r=1/(k+1)1−1/(k+1)=1/(k+1)(k+1−1)/(k+1)=1kS = \frac{a}{1-r} = \frac{1/(k+1)}{1 - 1/(k+1)} = \frac{1/(k+1)}{(k+1-1)/(k+1)} = \frac{1}{k}. The total sum of the original series is 2S=2k2S = \frac{2}{k}. We are given that this sum is 1/31/3. Therefore, 2k=13\frac{2}{k} = \frac{1}{3}, which implies k=6k = 6. For convergence, we need ∣r∣<1|r|<1, so ∣1k+1∣<1|\frac{1}{k+1}|<1, which means k>0k>0 or k<−2k<-2. Our solution k=6k=6 satisfies this condition.

Question 3

For what values of xx does the series ∑n=1∞(x−2)n3n−1\sum_{n=1}^{\infty} \frac{(x-2)^n}{3^{n-1}} converge?

  1. ∣x−2∣<1|x-2| < 1, which gives the interval (1,3)(1, 3)
  2. ∣x−2∣<13|x-2| < \frac{1}{3}, which gives the interval (53,73)\left(\frac{5}{3}, \frac{7}{3}\right)
  3. ∣x−2∣≤3|x-2| \leq 3, which gives the interval [−1,5][-1, 5]
  4. ∣x−2∣<3|x-2| < 3, which gives the interval (−1,5)(-1, 5) (correct answer)
Explanation: When you encounter a power series like this one, you're looking at a problem about radius and interval of convergence. The key is recognizing this as a geometric series and applying the convergence test. First, rewrite the series by factoring out constants: ∑n=1∞(x−2)n3n−1=∑n=1∞3⋅(x−2)n3n=3∑n=1∞(x−23)n\sum_{n=1}^{\infty} \frac{(x-2)^n}{3^{n-1}} = \sum_{n=1}^{\infty} \frac{3 \cdot (x-2)^n}{3^n} = 3\sum_{n=1}^{\infty} \left(\frac{x-2}{3}\right)^n This is a geometric series with first term a=x−23a = \frac{x-2}{3} and common ratio r=x−23r = \frac{x-2}{3}. A geometric series ∑arn−1\sum ar^{n-1} converges when ∣r∣<1|r| < 1. Therefore, convergence requires ∣x−23∣<1\left|\frac{x-2}{3}\right| < 1, which gives us ∣x−2∣<3|x-2| < 3. Solving this inequality: −3<x−2<3-3 < x-2 < 3, so −1<x<5-1 < x < 5, giving the interval (−1,5)(-1, 5). Choice A incorrectly uses ∣x−2∣<1|x-2| < 1, likely from forgetting to account for the denominator 3n−13^{n-1}. Choice B uses ∣x−2∣<13|x-2| < \frac{1}{3}, which appears to come from mishandling the algebraic manipulation of the geometric series form. Choice C includes the endpoints with ∣x−2∣≤3|x-2| \leq 3, but geometric series convergence is strict inequality—the series diverges exactly when ∣r∣=1|r| = 1. The correct answer is D: ∣x−2∣<3|x-2| < 3 gives (−1,5)(-1, 5). Study tip: Always rewrite power series in standard geometric form ∑arn\sum ar^n first, then apply ∣r∣<1|r| < 1 for convergence. Remember that geometric series convergence never includes the boundary points.

Question 4

What is the sum of the series ∑n=1∞(1en−1en+1)\sum_{n=1}^{\infty} (\frac{1}{e^n} - \frac{1}{e^{n+1}})?

  1. 1e\frac{1}{e} (correct answer)
  2. 11
  3. 1e−1\frac{1}{e-1}
  4. 0
Explanation: This is a telescoping series. The kk-th partial sum is Sk=(1e−1e2)+(1e2−1e3)+⋯+(1ek−1ek+1)=1e−1ek+1S_k = (\frac{1}{e} - \frac{1}{e^2}) + (\frac{1}{e^2} - \frac{1}{e^3}) + \cdots + (\frac{1}{e^k} - \frac{1}{e^{k+1}}) = \frac{1}{e} - \frac{1}{e^{k+1}}. As k→∞k \to \infty, 1ek+1→0\frac{1}{e^{k+1}} \to 0, so the sum is 1e\frac{1}{e}. Alternatively, splitting into geometric series: ∑n=1∞(1e)n−∑n=1∞(1e)n+1=1/e1−1/e−1/e21−1/e=1e−1−1e(e−1)=e−1e(e−1)=1e\sum_{n=1}^{\infty} (\frac{1}{e})^n - \sum_{n=1}^{\infty} (\frac{1}{e})^{n+1} = \frac{1/e}{1-1/e} - \frac{1/e^2}{1-1/e} = \frac{1}{e-1} - \frac{1}{e(e-1)} = \frac{e-1}{e(e-1)} = \frac{1}{e}.

Question 5

Let f(x)=2x3f(x) = \frac{2x}{3}. Define a sequence xnx_n by x0=9x_0=9 and xn+1=f(xn)x_{n+1} = f(x_n) for n≥0n \ge 0. Find the value of ∑n=0∞xn\sum_{n=0}^{\infty} x_n.

  1. 18
  2. 19
  3. 27 (correct answer)
  4. The series diverges.
Explanation: We can find the first few terms of the sequence to identify the pattern. x0=9x_0 = 9. x1=f(x0)=f(9)=2(9)3=6x_1 = f(x_0) = f(9) = \frac{2(9)}{3} = 6. x2=f(x1)=f(6)=2(6)3=4x_2 = f(x_1) = f(6) = \frac{2(6)}{3} = 4. The sequence is 9,6,4,…9, 6, 4, \ldots. This is a geometric sequence with first term a=9a = 9 and common ratio r=69=23r = \frac{6}{9} = \frac{2}{3}. The sum of the corresponding infinite series is S=a1−r=91−2/3=91/3=27S = \frac{a}{1-r} = \frac{9}{1 - 2/3} = \frac{9}{1/3} = 27.

Question 6

A ball is dropped from a height of 10 meters. Each time it bounces, it reaches a height that is 3/43/4 of the previous height. What is the total vertical distance traveled by the ball before it comes to rest?

  1. 30 m
  2. 40 m
  3. 70 m (correct answer)
  4. 80 m
Explanation: The total distance is the initial drop plus the sum of the distances of all the bounces (up and down). Initial drop: 10 m. First bounce (up and down): 10(34)+10(34)=20(34)10(\frac{3}{4}) + 10(\frac{3}{4}) = 20(\frac{3}{4}). Second bounce: 10(34)2+10(34)2=20(34)210(\frac{3}{4})^2 + 10(\frac{3}{4})^2 = 20(\frac{3}{4})^2. The total distance of the bounces is an infinite geometric series: ∑n=1∞20(34)n\sum_{n=1}^{\infty} 20(\frac{3}{4})^n. The first term is a=20(3/4)=15a = 20(3/4) = 15 and the ratio is r=3/4r = 3/4. The sum of the bounces is Sbounces=a1−r=151−3/4=151/4=60S_{bounces} = \frac{a}{1-r} = \frac{15}{1 - 3/4} = \frac{15}{1/4} = 60 m. The total vertical distance is the initial drop plus the sum of the bounce distances: 10+60=7010 + 60 = 70 m.

Question 7

A quantity yy changes over discrete time steps n=0,1,2,…n=0, 1, 2, \ldots such that yn+1=yn−14yny_{n+1} = y_n - \frac{1}{4}y_n, with y0=100y_0=100. What is the total sum of all values of yy, ∑n=0∞yn\sum_{n=0}^{\infty} y_n?

  1. 80
  2. 4003\frac{400}{3}
  3. 400 (correct answer)
  4. The sum does not converge.
Explanation: The recurrence relation is yn+1=yn−14yn=(1−14)yn=34yny_{n+1} = y_n - \frac{1}{4}y_n = (1 - \frac{1}{4})y_n = \frac{3}{4}y_n. This shows that the sequence {yn}\{y_n\} is a geometric sequence. The first term is a=y0=100a = y_0 = 100, and the common ratio is r=3/4r = 3/4. We need to find the sum of the corresponding infinite series, ∑n=0∞100(34)n\sum_{n=0}^{\infty} 100(\frac{3}{4})^n. Since ∣r∣=3/4<1|r| = 3/4 < 1, the series converges. The sum is S=a1−r=1001−3/4=1001/4=400S = \frac{a}{1-r} = \frac{100}{1 - 3/4} = \frac{100}{1/4} = 400.

Question 8

A square has a side length of 4. A second square is inscribed by connecting the midpoints of the sides of the first square. A third square is inscribed in the second square in the same way, and this process continues indefinitely. What is the sum of the areas of all the squares?

  1. 16
  2. 24
  3. 32 (correct answer)
  4. The sum is infinite.
Explanation: The first square has area A1=42=16A_1 = 4^2 = 16. When a new square is inscribed by connecting the midpoints, its side length s2s_2 can be found using the Pythagorean theorem. A corner triangle has legs of length 4/2=24/2 = 2. So, s22=22+22=8s_2^2 = 2^2 + 2^2 = 8. The area of the second square is A2=s22=8A_2 = s_2^2 = 8. The ratio of the areas is A2/A1=8/16=1/2A_2/A_1 = 8/16 = 1/2. This ratio holds for all subsequent squares. The sum of the areas is an infinite geometric series: 16+8+4+⋯16 + 8 + 4 + \cdots. The first term is a=16a=16 and the common ratio is r=1/2r=1/2. The sum is S=a1−r=161−1/2=161/2=32S = \frac{a}{1-r} = \frac{16}{1-1/2} = \frac{16}{1/2} = 32.

Question 9

A ball is dropped from a height of 64 feet. After each bounce, it reaches a height that is 34\frac{3}{4} of its previous height. What is the total vertical distance traveled by the ball when it comes to rest?

  1. 192 feet, calculated as the sum of downward distances only
  2. 320 feet, accounting for both upward and downward motion
  3. 448 feet, including the initial drop and all subsequent motion (correct answer)
  4. 256 feet, using only the geometric series for upward motion
Explanation: The ball travels 64 feet down initially. Then it bounces to height 64⋅34=4864 \cdot \frac{3}{4} = 48 feet (48 up, 48 down), then 48⋅34=3648 \cdot \frac{3}{4} = 36 feet (36 up, 36 down), etc. Total distance = 64+2(48+36+27+…)=64+2⋅48∑k=0∞(34)k=64+96⋅11−34=64+96⋅4=64+384=44864 + 2(48 + 36 + 27 + \ldots) = 64 + 2 \cdot 48 \sum_{k=0}^{\infty} \left(\frac{3}{4}\right)^k = 64 + 96 \cdot \frac{1}{1-\frac{3}{4}} = 64 + 96 \cdot 4 = 64 + 384 = 448 feet. Choice A only counts downward motion. Choice B forgets the initial drop. Choice D only counts upward bounces.

Question 10

Evaluate the sum ∑n=2∞5⋅3n−24n−1\sum_{n=2}^{\infty} \frac{5 \cdot 3^{n-2}}{4^{n-1}}.

  1. 5 (correct answer)
  2. 203\frac{20}{3}
  3. 809\frac{80}{9}
  4. The series diverges.
Explanation: First, rewrite the general term of the series to fit the form arnar^n. 5⋅3n−24n−1=5⋅3n⋅3−24n⋅4−1=5⋅(1/9)(1/4)(34)n=59⋅4⋅(34)n=209(34)n\frac{5 \cdot 3^{n-2}}{4^{n-1}} = \frac{5 \cdot 3^n \cdot 3^{-2}}{4^n \cdot 4^{-1}} = \frac{5 \cdot (1/9)}{(1/4)} (\frac{3}{4})^n = \frac{5}{9} \cdot 4 \cdot (\frac{3}{4})^n = \frac{20}{9} (\frac{3}{4})^n. The series is 209∑n=2∞(34)n\frac{20}{9} \sum_{n=2}^{\infty} (\frac{3}{4})^n. This is a geometric series with ratio r=3/4r = 3/4. The sum starts at n=2n=2, so the first term of ∑n=2∞(34)n\sum_{n=2}^{\infty} (\frac{3}{4})^n is (3/4)2=9/16(3/4)^2 = 9/16. The sum of this part is S=a1−r=9/161−3/4=9/161/4=94S = \frac{a}{1-r} = \frac{9/16}{1 - 3/4} = \frac{9/16}{1/4} = \frac{9}{4}. The total sum is 209⋅S=209⋅94=5\frac{20}{9} \cdot S = \frac{20}{9} \cdot \frac{9}{4} = 5.

Question 11

Which of the following describes the set of all cc for which the series ∑n=1∞(11+c2)n\sum_{n=1}^{\infty} (\frac{1}{1+c^2})^n converges?

  1. c=0c = 0
  2. c>0c > 0
  3. c≠0c \ne 0 (correct answer)
  4. All real numbers cc.
Explanation: This is a geometric series with ratio r=11+c2r = \frac{1}{1+c^2}. For the series to converge, we need ∣r∣<1|r| < 1. So, we must satisfy ∣11+c2∣<1|\frac{1}{1+c^2}| < 1. Since c2≥0c^2 \ge 0, the denominator 1+c21+c^2 is always greater than or equal to 1. This means 0<11+c2≤10 < \frac{1}{1+c^2} \le 1. The convergence condition ∣r∣<1|r|<1 requires a strict inequality. Therefore, we need 11+c2<1\frac{1}{1+c^2} < 1. This is true as long as the denominator 1+c2>11+c^2 > 1, which means c2>0c^2 > 0. This condition holds for all real numbers cc except for c=0c=0. If c=0c=0, the ratio r=1r=1, and the series diverges. Thus, the series converges for all c≠0c \ne 0.

Question 12

Let f(x)=∑n=1∞xn2nf(x) = \sum_{n=1}^{\infty} \frac{x^n}{2^n}. What is the value of the derivative f′(1)f'(1)?

  1. 1
  2. 2 (correct answer)
  3. ln⁡(2)\ln(2)
  4. The derivative does not exist.
Explanation: The function f(x)f(x) is a geometric series ∑n=1∞(x2)n\sum_{n=1}^{\infty} (\frac{x}{2})^n. This series converges for ∣x/2∣<1|x/2|<1, i.e., ∣x∣<2|x|<2. The sum is f(x)=first term1−ratio=x/21−x/2=x2−xf(x) = \frac{\text{first term}}{1-\text{ratio}} = \frac{x/2}{1-x/2} = \frac{x}{2-x}. We can find the derivative of this function using the quotient rule: f′(x)=(1)(2−x)−(x)(−1)(2−x)2=2−x+x(2−x)2=2(2−x)2f'(x) = \frac{(1)(2-x) - (x)(-1)}{(2-x)^2} = \frac{2-x+x}{(2-x)^2} = \frac{2}{(2-x)^2}. Now, we evaluate the derivative at x=1x=1: f′(1)=2(2−1)2=21=2f'(1) = \frac{2}{(2-1)^2} = \frac{2}{1} = 2. Since x=1x=1 is within the interval of convergence, this is a valid operation.

Question 13

The number 2.135‾2.1\overline{35} can be expressed as a fraction pq\frac{p}{q} in lowest terms. What is the value of p+qp+q?

  1. 1552 (correct answer)
  2. 3104
  3. 249
  4. 136
Explanation: The number can be written as 2.1+0.035‾2.1 + 0.0\overline{35}. The repeating part, 0.0353535...0.0353535..., can be expressed as an infinite geometric series: 351000+35100000+⋯\frac{35}{1000} + \frac{35}{100000} + \cdots. This series has a first term a=351000a = \frac{35}{1000} and a common ratio r=1100r = \frac{1}{100}. Its sum is S=a1−r=35/10001−1/100=35/100099/100=35990S = \frac{a}{1-r} = \frac{35/1000}{1 - 1/100} = \frac{35/1000}{99/100} = \frac{35}{990}. The original number is 2.1+35990=2110+35990=21⋅99+35990=2079+35990=21149902.1 + \frac{35}{990} = \frac{21}{10} + \frac{35}{990} = \frac{21 \cdot 99 + 35}{990} = \frac{2079 + 35}{990} = \frac{2114}{990}. Simplifying this fraction by dividing the numerator and denominator by 2 gives 1057495\frac{1057}{495}. This is in lowest terms, so p=1057p=1057 and q=495q=495. Their sum is p+q=1057+495=1552p+q = 1057 + 495 = 1552.

Question 14

Consider the series ∑n=1∞(ln⁡x)2n\sum_{n=1}^{\infty} (\ln x)^{2n}. If the sum of this series is 1/81/8, what is a possible value of xx?

  1. e3e^3
  2. e1/3e^{1/3} (correct answer)
  3. e−3e^{-3}
  4. e−1/3e^{-1/3}
Explanation: The series can be written as ∑n=1∞((ln⁡x)2)n\sum_{n=1}^{\infty} ((\ln x)^2)^n. This is a geometric series with ratio r=(ln⁡x)2r = (\ln x)^2. The series starts at n=1n=1, so the first term is a=(ln⁡x)2a = (\ln x)^2. The sum is S=a1−r=(ln⁡x)21−(ln⁡x)2S = \frac{a}{1-r} = \frac{(\ln x)^2}{1-(\ln x)^2}. We are given that the sum is 1/81/8. So, (ln⁡x)21−(ln⁡x)2=18\frac{(\ln x)^2}{1-(\ln x)^2} = \frac{1}{8}. Let y=(ln⁡x)2y = (\ln x)^2. Then y1−y=18\frac{y}{1-y} = \frac{1}{8}, which gives 8y=1−y⇒9y=1⇒y=1/98y = 1-y \Rightarrow 9y=1 \Rightarrow y=1/9. So, (ln⁡x)2=1/9(\ln x)^2 = 1/9. This implies ln⁡x=±1/3\ln x = \pm 1/3. The possible values for xx are x=e1/3x = e^{1/3} or x=e−1/3x = e^{-1/3}. Both are valid choices as they lead to a ratio r=1/9r=1/9 which satisfies ∣r∣<1|r|<1. From the options, e1/3e^{1/3} is a possible value.

Question 15

An infinite geometric series has a first term of 12 and a sum of 16. What is the sum of the first 3 terms of this series?

  1. 14\frac{1}{4}
  2. 394\frac{39}{4}
  3. 634\frac{63}{4} (correct answer)
  4. 15
Explanation: First, find the common ratio rr using the formula for the sum of an infinite geometric series, S=a1−rS = \frac{a}{1-r}. We have 16=121−r16 = \frac{12}{1-r}. Solving for rr gives 16(1−r)=12⇒1−r=1216=34⇒r=1416(1-r) = 12 \Rightarrow 1-r = \frac{12}{16} = \frac{3}{4} \Rightarrow r = \frac{1}{4}. The sum of the first nn terms is given by Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r}. For the first 3 terms, S3=12(1−(1/4)3)1−1/4=12(1−1/64)3/4=12(63/64)3/4=16⋅6364=634S_3 = \frac{12(1 - (1/4)^3)}{1 - 1/4} = \frac{12(1 - 1/64)}{3/4} = \frac{12(63/64)}{3/4} = 16 \cdot \frac{63}{64} = \frac{63}{4}. Alternatively, the first three terms are 1212, 12(14)=312(\frac{1}{4})=3, and 3(14)=343(\frac{1}{4})=\frac{3}{4}. Their sum is 12+3+34=15+34=63412 + 3 + \frac{3}{4} = 15 + \frac{3}{4} = \frac{63}{4}.

Question 16

What is the sum of the series ∑n=1∞4n−3n−15n\sum_{n=1}^{\infty} \frac{4^n - 3^{n-1}}{5^n}?

  1. 72\frac{7}{2} (correct answer)
  2. 172\frac{17}{2}
  3. 196\frac{19}{6}
  4. The series diverges.
Explanation: The series can be split into two convergent geometric series: ∑n=1∞4n5n−∑n=1∞3n−15n\sum_{n=1}^{\infty} \frac{4^n}{5^n} - \sum_{n=1}^{\infty} \frac{3^{n-1}}{5^n}. The first series is ∑n=1∞(45)n\sum_{n=1}^{\infty} (\frac{4}{5})^n, with first term a1=4/5a_1 = 4/5 and ratio r1=4/5r_1 = 4/5. Its sum is S1=4/51−4/5=4S_1 = \frac{4/5}{1 - 4/5} = 4. The second series term can be rewritten as 3n−15n=133n5n=13(35)n\frac{3^{n-1}}{5^n} = \frac{1}{3} \frac{3^n}{5^n} = \frac{1}{3}(\frac{3}{5})^n. This series has first term (at n=1) a2=13(35)=15a_2 = \frac{1}{3}(\frac{3}{5}) = \frac{1}{5} and ratio r2=3/5r_2 = 3/5. Its sum is S2=1/51−3/5=1/52/5=12S_2 = \frac{1/5}{1 - 3/5} = \frac{1/5}{2/5} = \frac{1}{2}. The total sum is S1−S2=4−12=72S_1 - S_2 = 4 - \frac{1}{2} = \frac{7}{2}.

Question 17

Let S1=∑n=0∞(x−2)nS_1 = \sum_{n=0}^{\infty} (x-2)^n and S2=∑n=0∞(x3)nS_2 = \sum_{n=0}^{\infty} (\frac{x}{3})^n. For which open interval of xx values do both series converge?

  1. (−3,3)(-3, 3)
  2. (1,3)(1, 3) (correct answer)
  3. (1,∞)(1, \infty)
  4. (−3,1)(-3, 1)
Explanation: For a geometric series to converge, the absolute value of its common ratio must be less than 1. For S1S_1, the ratio is r1=x−2r_1 = x-2. Convergence requires ∣x−2∣<1|x-2| < 1, which is equivalent to −1<x−2<1-1 < x-2 < 1. Adding 2 to all parts gives 1<x<31 < x < 3. For S2S_2, the ratio is r2=x/3r_2 = x/3. Convergence requires ∣x/3∣<1|x/3| < 1, which is equivalent to −1<x/3<1-1 < x/3 < 1. Multiplying by 3 gives −3<x<3-3 < x < 3. For both series to converge, xx must be in the intersection of these two intervals: (1,3)∩(−3,3)(1, 3) \cap (-3, 3). The intersection is the interval (1,3)(1, 3).

Question 18

Let an=∫02(x/3)ndxa_n = \int_0^2 (x/3)^n dx. Evaluate ∑n=0∞an\sum_{n=0}^{\infty} a_n.

  1. 3ln⁡(3)3\ln(3) (correct answer)
  2. ln⁡(3)\ln(3)
  3. 2
  4. The series diverges.
Explanation: We can swap the summation and integration because the series converges uniformly on the interval of integration. ∑n=0∞an=∑n=0∞∫02(x/3)ndx=∫02∑n=0∞(x/3)ndx\sum_{n=0}^{\infty} a_n = \sum_{n=0}^{\infty} \int_0^2 (x/3)^n dx = \int_0^2 \sum_{n=0}^{\infty} (x/3)^n dx. The expression inside the integral is a geometric series with ratio r=x/3r = x/3. For x∈[0,2]x \in [0, 2], ∣r∣≤2/3<1|r| \le 2/3 < 1, so the series converges to 11−x/3=33−x\frac{1}{1 - x/3} = \frac{3}{3-x}. Now, we evaluate the integral: ∫0233−xdx\int_0^2 \frac{3}{3-x} dx. Using a u-substitution with u=3−xu=3-x and du=−dxdu=-dx, the integral becomes ∫313−du=−3∫311udu=3∫131udu=3[ln⁡∣u∣]13=3(ln⁡(3)−ln⁡(1))=3ln⁡(3)\int_3^1 \frac{3}{-du} = -3 \int_3^1 \frac{1}{u} du = 3 \int_1^3 \frac{1}{u} du = 3[\ln|u|]_1^3 = 3(\ln(3) - \ln(1)) = 3\ln(3).

Question 19

A geometric series is defined by ∑n=2∞3⋅2n+15n\sum_{n=2}^{\infty} \frac{3 \cdot 2^{n+1}}{5^n}. What is the sum of this series?

  1. 125\frac{12}{5}
  2. 85\frac{8}{5} (correct answer)
  3. 4
  4. 10
Explanation: The terms of the series can be rewritten to identify the common ratio. 3⋅2n+15n=3⋅2⋅2n5n=6(25)n\frac{3 \cdot 2^{n+1}}{5^n} = \frac{3 \cdot 2 \cdot 2^n}{5^n} = 6(\frac{2}{5})^n. This is a geometric series with common ratio r=2/5r = 2/5. Since ∣r∣<1|r| < 1, the series converges. The summation starts at n=2n=2, so the first term aa is 6(25)2=6(425)=24256(\frac{2}{5})^2 = 6(\frac{4}{25}) = \frac{24}{25}. The sum of an infinite geometric series is S=a1−rS = \frac{a}{1-r}. Plugging in the values, S=24/251−2/5=24/253/5=2425⋅53=85S = \frac{24/25}{1 - 2/5} = \frac{24/25}{3/5} = \frac{24}{25} \cdot \frac{5}{3} = \frac{8}{5}.

Question 20

Let A=∑n=1∞(1x)nA = \sum_{n=1}^{\infty} (\frac{1}{x})^n and B=∑n=0∞(1x+1)nB = \sum_{n=0}^{\infty} (\frac{1}{x+1})^n. If A=BA=B and both series converge, what is the value of xx?

  1. 1−52\frac{1 - \sqrt{5}}{2}
  2. 1+52\frac{1 + \sqrt{5}}{2} (correct answer)
  3. 1
  4. There is no such value of x.
Explanation: First, find the sums of the series in terms of xx. For series A, a=1/xa = 1/x and r=1/xr = 1/x. The sum is A=1/x1−1/x=1x−1A = \frac{1/x}{1 - 1/x} = \frac{1}{x-1}. For A to converge, ∣1/x∣<1|1/x| < 1, which means ∣x∣>1|x| > 1. For series B, a=1a = 1 and r=1/(x+1)r = 1/(x+1). The sum is B=11−1/(x+1)=x+1xB = \frac{1}{1 - 1/(x+1)} = \frac{x+1}{x}. For B to converge, ∣1/(x+1)∣<1|1/(x+1)| < 1, which means ∣x+1∣>1|x+1| > 1. This implies x>0x > 0 or x<−2x < -2. The combined condition for convergence is x>1x > 1 or x<−2x < -2. Now, set A=BA=B: 1x−1=x+1x\frac{1}{x-1} = \frac{x+1}{x}. Cross-multiplying gives x=(x−1)(x+1)=x2−1x = (x-1)(x+1) = x^2 - 1. This leads to the quadratic equation x2−x−1=0x^2 - x - 1 = 0. Using the quadratic formula, x=1±1−4(1)(−1)2=1±52x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}. We must check which solution satisfies the convergence condition. x1=1+52≈1.618x_1 = \frac{1 + \sqrt{5}}{2} \approx 1.618, which is greater than 1, so it is a valid solution. x2=1−52≈−0.618x_2 = \frac{1 - \sqrt{5}}{2} \approx -0.618, which is not in the convergence interval. Thus, the only solution is x=1+52x = \frac{1 + \sqrt{5}}{2}.