← Back to the Series Binomial Theorem Limits worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: A-APR.C.5, A-SSE.B.4, F-BF.A.2 · Level: Grade 12 Precalculus
Problem 1 Evaluating a Sum
Substitute k = 1 to 6: 2 + 5 + 10 + 17 + 26 + 37 = 97.
Answer: 97
Problem 2 Writing Sigma Notation
The terms increase by 3, starting at 4, so the rule is 3k + 1 for k = 1 to 5. Check: k = 1 gives 4 and k = 5 gives 16.
Answer: ∑k=15 (3k + 1)
Problem 3 An Arithmetic Series
The difference is 4. The number of terms: n = 51 − 74 + 1 = 12. Sum = 122(7 + 51) = 6(58) = 348.
Answer: 348
Problem 4 A Geometric Series
a1 = 4 and r = 3. S7 = 4 · 1 − 3⁷1 − 3 = 4 · −2,186−2 = 4(1,093) = 4,372.
Answer: 4,372
Problem 5 An Infinite Sum
r = 69 = 23, and |r| < 1, so the series converges. S = 91 − 23 = 913 = 27.
Answer: 27
Problem 6 Divergent or Not?
The ratio is 9.6 ÷ 8 = 1.2. Since |1.2| ≥ 1, the terms grow and the series diverges, so it has no sum.
Answer: no sum; it diverges because r = 1.2
Problem 7 How Many Terms?
Sn = n2[2(−4) + 3(n − 1)] = n(3n − 11)2 = 217. So 3n² − 11n − 434 = 0.
The discriminant is 121 + 5,208 = 5,329 = 73², so n = 11 + 736 = 14 (the negative root is rejected).
Check: the 14th term is −4 + 13(3) = 35 and 142(−4 + 35) = 217.
Answer: n = 14
Problem 8 An Alternating Sum
This is geometric with a1 = 5 and r = −2, with 7 terms. S = 5 · 1 − (−2)⁷1 − (−2) = 5 · 1293 = 215.
Check: 5 − 10 + 20 − 40 + 80 − 160 + 320 = 215.
Answer: 215
Problem 9 A Repeating Decimal
0.272727… = 0.27 + 0.0027 + 0.000027 + …, with a1 = 0.27 and r = 0.01.
S = 0.271 − 0.01 = 0.270.99 = 2799 = 311.
Answer: 311
Problem 10 An Alternating Infinite Series
a1 = 6 and r = −12, and |r| < 1. S = 61 − (−12) = 632 = 4.
Answer: 4
Problem 11 Pascal’s Triangle Expansion
Row 5 is 1, 5, 10, 10, 5, 1. Then (x + 2)⁵ = x⁵ + 5(2)x⁴ + 10(4)x³ + 10(8)x² + 5(16)x + 32.
Answer: x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32
Problem 12 Expanding with a Coefficient
Use 1, 4, 6, 4, 1 with a = 2x and b = −3: (2x)⁴ + 4(2x)³(−3) + 6(2x)²(−3)² + 4(2x)(−3)³ + (−3)⁴.
= 16x⁴ − 96x³ + 216x² − 216x + 81.
Answer: 16x⁴ − 96x³ + 216x² − 216x + 81
Problem 13 A Single Coefficient
The term with x³ is C(7, 3)x³·4⁴. C(7, 3) = 7·6·53·2·1 = 35 and 4⁴ = 256.
The coefficient is 35 × 256 = 8,960.
Answer: 8,960
Problem 14 A Limit from a Table
(a) For x ≠ 3, f(x) = x + 3. So the values are 5.9, 5.99, 5.999, 6.001 and 6.01.
(b) The outputs approach 6 from both sides, so the limit is 6. At x = 3 the formula gives 0/0, so f(3) is undefined and the graph has a hole at (3, 6).
Answer: limit = 6; f(3) is undefined
Problem 15 Theater Seating
(a) Arithmetic with a1 = 14 and d = 2: a18 = 14 + 17(2) = 48.
(b) S18 = 182(14 + 48) = 9(62) = 558.
Answer: (a) 48 seats; (b) 558 seats
Problem 16 Monthly Savings Plan
The deposits form a geometric series with a1 = 40, r = 1.05 and n = 24.
S24 = 40 · 1.05²⁴ − 10.05 ≈ 40 · 3.2251 − 10.05 ≈ 40(44.502) ≈ $1,780.08.
Answer: about $1,780.08
Problem 17 Bouncing Ball
The first drop is 10 ft. After that, each bounce goes up and down a distance equal to 2 × the new height. The heights are 7, 4.9, 3.43, … , a geometric series with a1 = 7 and r = 0.7.
Total up-and-down distance = 2 · 71 − 0.7 = 2(23.33…) = 46.67 ft.
Total = 10 + 46.67 = 56.67 ft.
Answer: about 56.67 feet
Problem 18 Medicine Dosage
(a) After dose 1: 200 mg. After dose 2: 200 + 0.3(200) = 260 mg. After dose 3: 200 + 0.3(260) = 278 mg, which is 200 + 60 + 18.
(b) The amount is 200(1 + 0.3 + 0.3² + …), an infinite geometric series with r = 0.3. S = 2001 − 0.3 = 20007 ≈ 285.7 mg.
Answer: 200, 260, 278 mg; long-run amount about 285.7 mg
Problem 19 A Special Term
The general term is C(6, k)(x²)6−k(x⁻¹)k = C(6, k)x12−3k.
(a) The exponent is 0 when 12 − 3k = 0, so k = 4, and C(6, 4) = 15.
(b) The exponent is 3 when 12 − 3k = 3, so k = 3, and C(6, 3) = 20.
Answer: (a) 15; (b) 20
Problem 20 Limits of a Function and a Sequence
(a) x³ + 1 = (x + 1)(x² − x + 1). For x ≠ −1 the expression equals x² − x + 1, and at x = −1 that is 1 + 1 + 1 = 3.
(b) Divide the top and bottom by n²: 6 + 5/n²2 − 1/n + 3/n². As n grows, the small terms go to 0, leaving 62 = 3.
(c) A limit only depends on the values near −1, not at −1 itself. The factor x + 1 cancels for every x ≠ −1, so the outputs approach 3.
Answer: (a) 3; (b) 3; (c) the limit uses nearby values, not the value at −1
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