Series, Binomial Theorem and an Introduction to Limits: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Sequences Series Worksheet Answer Key: problems 1–5

Problem 1: Evaluate a sum

Substitute \(k=1,2,3,4,5\) into \(3k-1\): 2, 5, 8, 11, 14.

\(2+5+8+11+14=\)40

Problem 2: Sums of powers

1. \(1+4+9+16=\)30
2. \(1+2+4+8=\)15

Problem 3: Write in sigma notation

1. Each term is \(3k\) for \(k=1\) to 5: \(\sum_{k=1}^{5}3k\)
2. Each term is \(2^{k}\) for \(k=1\) to 5: \(\sum_{k=1}^{5}2^{k}\)

Problem 4: Explicit and recursive rules

1. Common difference 4. Explicit: \(a_{n}=5+4(n-1)=4n+1\). Recursive: \(a_{1}=5\), \(a_{n}=a_{n-1}+4\). \(a_{20}=4(20)+1=\)81
2. Common ratio 3. Explicit: \(a_{n}=2\cdot 3^{n-1}\). Recursive: \(a_{1}=2\), \(a_{n}=3a_{n-1}\). \(a_{6}=2\cdot 3^{5}=2\cdot 243=\)486

Problem 5: Find the nth term

1. \(a_{20}=7+4(19)=7+76=\)83
2. \(7+4(n-1)=103\Rightarrow 4(n-1)=96\Rightarrow n-1=24\), so term 25

Precalculus Sequences Series Worksheet Answer Key: problems 6–10

Problem 6: Arithmetic series

\(d=7\), so \(a_{15}=12+7(14)=110\).

\(S_{15}=\frac{15(12+110)}{2}=15\cdot 61=\)915

Problem 7: Decreasing series

\(d=-3\). Solve \(100-3(n-1)=1\): \(n-1=33\), so \(n=34\) terms.

\(S_{34}=\frac{34(100+1)}{2}=17\cdot 101=\)1,717

Problem 8: Finite geometric series

1. \(r=2\): \(S_{8}=\frac{3(1-2^{8})}{1-2}=\frac{3(-255)}{-1}=\)765
2. \(r=\frac{1}{3}\), \(n=5\): \(S_{5}=\frac{81(1-\frac{1}{243})}{\frac{2}{3}}=81\cdot\frac{242}{243}\cdot\frac{3}{2}=\)121

Problem 9: Does it converge?

1. \(r=\frac{1}{2}\), \(|r| < 1\): converges to \(\frac{8}{1-1/2}=\)16
2. \(r=2\): diverges
3. \(r=-\frac{1}{3}\): converges to \(\frac{9}{1+1/3}=\frac{9}{4/3}=\)27/4
4. \(r=1.1\): diverges

Problem 10: Infinite geometric sum

1. \(r=\frac{1}{3}\): \(\frac{12}{1-1/3}=12\cdot\frac{3}{2}=\)18
2. \(\frac{5}{1-0.2}=\frac{5}{0.8}=\)6.25

Precalculus Sequences Series Worksheet Answer Key: problems 11–15

Problem 11: Repeating decimals

1. \(0.27+0.0027+\) … has \(a_{1}=0.27\), \(r=0.01\): \(\frac{0.27}{0.99}=\frac{27}{99}=\)3/11
2. \(a_{1}=0.6\), \(r=0.1\): \(\frac{0.6}{0.9}=\)2/3

Problem 12: Pascal's triangle

Each entry is the sum of the two above it: row 5 is 1, 5, 10, 10, 5, 1 and row 6 is 1, 6, 15, 20, 15, 6, 1.

1. The missing entries are 10, 15 and 20.
2. Row 7: 1, 7, 21, 35, 35, 21, 7, 1

Problem 13: Expand a binomial

1. Row 5 is 1, 5, 10, 10, 5, 1: \(x^{5}+5\cdot 2x^{4}+10\cdot 4x^{3}+10\cdot 8x^{2}+5\cdot 16x+32=\)\(x^{5}+10x^{4}+40x^{3}+80x^{2}+80x+32\)
2. Row 3 is 1, 3, 3, 1 with \(b=-3\): \(x^{3}+3(-3)x^{2}+3(9)x+(-27)=\)\(x^{3}-9x^{2}+27x-27\)

Problem 14: Binomial with a coefficient

Row 4 is 1, 4, 6, 4, 1 with \(a=2x\) and \(b=-1\).

\((2x)^{4}+4(2x)^{3}(-1)+6(2x)^{2}(1)+4(2x)(-1)+1=\)\(16x^{4}-32x^{3}+24x^{2}-8x+1\)

Problem 15: Find a specific term

1. The \(x^{3}\) term is \(C(7,3)\,x^{3}\,3^{4}=35\cdot 81\,x^{3}\). The coefficient is 2,835.
2. \(C(6,2)(2x)^{2}(1)^{4}=15\cdot 4x^{2}=\)\(60x^{2}\)

Precalculus Sequences Series Worksheet Answer Key: problems 16–20

Problem 16: Limit of a sequence

1. \(a_{10}=\frac{31}{12}\approx 2.5833\); \(a_{100}=\frac{301}{102}\approx 2.9510\); \(a_{1000}=\frac{3001}{1002}\approx 2.9950\)
2. Divide top and bottom by \(n\): \(\frac{3+1/n}{1+2/n}\to\frac{3}{1}\). The limit is 3.

Problem 17: Limit from a graph

1. No. Substituting \(x=3\) gives \(\frac{0}{0}\), which is why the graph has an open circle.
2. Values near 3 approach the height of the hole: 6.

Problem 18: Limit by factoring

1. \(\frac{(x+4)(x-3)}{x-3}=x+4\) for \(x\neq 3\), so the limit is 7.
2. \(\frac{(x+5)(x-5)}{x-5}=x+5\) for \(x\neq 5\), so the limit is 10.

Problem 19: When does it converge?

1. The ratio is \(\frac{x}{3}\). Convergence needs \(|\frac{x}{3}| < 1\), so \(-3 < x < 3\).
2. \(r=\frac{1}{3}\): \(\frac{1}{1-1/3}=\)3/2
3. \(r=-\frac{1}{2}\): \(\frac{1}{1+1/2}=\)2/3

Problem 20: Theater seats

The seats rise by 3 each row, so \(a_{n}=4+3(n-1)\).

Rows 1 to 6: \(S_{6}=\frac{6(4+19)}{2}=\)69 seats.

Row 20 has \(a_{20}=4+57=61\) seats, so \(S_{20}=\frac{20(4+61)}{2}=\)650 seats.

Precalculus Sequences Series Worksheet Answer Key: problems 21–25

Problem 21: Doubling savings

Geometric with \(a_{1}=0.01\), \(r=2\), \(n=12\).

\(S_{12}=\frac{0.01(1-2^{12})}{1-2}=0.01(4095)=\)$40.95

Problem 22: Salary raises

Recursive: \(a_{1}=40000\), \(a_{n}=a_{n-1}+2500\).

Year 10: \(a_{10}=40000+9(2500)=\)$62,500.

Total: \(S_{10}=\frac{10(40000+62500)}{2}=\)$512,500.

Problem 23: Bouncing ball

Down 10 ft first. After that each bounce goes up and down: heights 6, 3.6, 2.16, … form a geometric series with \(a_{1}=6\), \(r=0.6\).

Up-and-down total: \(2\cdot\frac{6}{1-0.6}=2\cdot 15=30\).

Distance: \(10+30=\)40 feet.

Problem 24: Medicine in the body

Day 1: 100 mg. Day 2: \(100+0.2(100)=120\) mg. Day 3: \(100+0.2(120)=124\) mg, so 124 mg.

Day \(n\) is \(100(1+0.2+0.2^{2}+\) … \()\). The long-run total is \(\frac{100}{1-0.2}=\)125 mg.

Problem 25: Cleaning a pond

The points level off near 9 pounds.

Exactly: \(a_{1}=6\), \(r=\frac{1}{3}\), so the total is \(\frac{6}{1-1/3}=6\cdot\frac{3}{2}=\)9 pounds.

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