Sequences, Series, and Induction: practice solutions, Grade 12 – download the PDF

1 Terms of an arithmetic sequence ★★★

Add 6 each time: 4, 10, 16, 22, 28.

\(a_{12} = a_1 + 11d = 4 + 11 \times 6 = 70\).

Answer: the first five terms are 4, 10, 16, 22, 28 and \(a_{12} = 70\).

2 Arithmetic, geometric, or neither? ★★★

(a) Ratios: \(9/3 = 27/9 = 81/27 = 3\), so it is geometric with \(r = 3\).

(b) Differences: \(-7\) each time, so it is arithmetic with \(d = -7\).

(c) Differences are 3, 5, 7 (not constant) and ratios \(5/2\), \(10/5 = 2\) (not constant), so it is neither.

3 Expand a sum ★★★

Replace \(k\) by 1, 2, 3, 4, 5: \(1 + 4 + 7 + 10 + 13 = 35\).

Answer: 35.

4 Fifth term of a geometric sequence ★★★

\(a_5 = a_1 r^{4} = 2 \times 5^4 = 2 \times 625 = 1250\).

Answer: \(a_5 = 1250\).

5 Does the series have a sum? ★★★

(a) False. The ratio is \(r = 2\), and \(|r| \ge 1\), so the series diverges.

(b) True. The ratio is \(r = 0.2\) with \(|r| \lt 1\). The sum is \(\dfrac{5}{1 - 0.2} = \dfrac{5}{0.8} = 6.25\).

6 Recursive to explicit ★★★

Terms: 5, 13, 21, 29. Each term is 8 more than the last, so the sequence is arithmetic with \(d = 8\).

\(a_n = 5 + (n-1)\cdot 8 = 8n - 3\). Check: \(a_4 = 32 - 3 = 29\).

7 Expand with Pascal’s triangle ★★★

Row 5 is 1, 5, 10, 10, 5, 1. Since \(1\) raised to any power is 1:

\((x+1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1\).

Check with \(x = 1\): \(2^5 = 32 = 1+5+10+10+5+1\).

8 Stadium seating ★★★

This is arithmetic with \(a_1 = 22\), \(d = 3\), \(n = 18\).

\(a_{18} = 22 + 17\times 3 = 73\).

\(S_{18} = \dfrac{18(22 + 73)}{2} = 9 \times 95 = 855\).

Answer: the last row has 73 seats, and the section has 855 seats.

9 Weekly savings ★★★

Arithmetic with \(a_1 = 50\), \(d = 5\), \(n = 26\).

\(a_{26} = 50 + 25 \times 5 = 175\).

\(S_{26} = \dfrac{26(50 + 175)}{2} = 13 \times 225 = 2925\).

Answer: 175 dollars in week 26 and 2,925 dollars in total.

10 A spreading rumor ★★★

The daily counts 6, 12, 24, ... form a geometric sequence with \(a_1 = 6\), \(r = 2\).

\(S_{10} = 6\cdot\dfrac{2^{10} - 1}{2 - 1} = 6 \times 1023 = 6138\).

Answer: 6,138 students (a large school indeed!).

11 How many terms? ★★★

(a) \(r = 4\). Solve \(3\cdot 4^{n-1} = 3072\): \(4^{n-1} = 1024 = 4^5\), so \(n - 1 = 5\) and \(n = 6\).

(b) \(S_6 = 3\cdot\dfrac{4^6 - 1}{4 - 1} = 4096 - 1 = 4095\).

Check: \(3 + 12 + 48 + 192 + 768 + 3072 = 4095\).

12 Repeating decimal as a fraction ★★★

\(0.4545\ldots = 0.45 + 0.0045 + 0.000045 + \cdots\), so \(a_1 = 0.45\) and \(r = 0.01\).

\(S = \dfrac{0.45}{1 - 0.01} = \dfrac{0.45}{0.99} = \dfrac{45}{99} = \dfrac{5}{11}\).

Check: \(5 \div 11 = 0.4545\ldots\)

13 Using the sum rules ★★★

(a) \(3\cdot\dfrac{40\cdot 41}{2} + 2\cdot 40 = 3 \times 820 + 80 = 2540\).

(b) \(\dfrac{12 \cdot 13 \cdot 25}{6} = 650\).

14 A Fibonacci-style sequence ★★★

Terms: 1, 3, 4, 7, 11, 18, 29, 47, 76.

So \(f_9 = 76\). Since \(f_8 = 47 \le 50\) and \(f_9 = 76 \gt 50\), the first index is \(n = 9\).

15 Two coefficients in a binomial expansion ★★★

(a) The general term is \(\dbinom{6}{k}(2x)^{k}(-1)^{6-k}\). For \(x^3\): \(k = 3\), giving \(20 \cdot 8 \cdot (-1)^3 = -160\).

(b) The general term is \(\dbinom{6}{k}(x^2)^{6-k}(x^{-1})^{k} = \dbinom{6}{k}x^{12 - 3k}\). The exponent is 0 when \(k = 4\), so the constant term is \(\dbinom{6}{4} = 15\).

16 Induction: a sum of products ★★★

Base case. \(n = 1\): left side \(1\cdot 2 = 2\), right side \(\dfrac{1\cdot 2\cdot 3}{3} = 2\).

Inductive step. Assume the formula for \(n = k\). Then the sum up to \(k+1\) is \[\dfrac{k(k+1)(k+2)}{3} + (k+1)(k+2) = (k+1)(k+2)\left(\dfrac{k}{3} + 1\right) = \dfrac{(k+1)(k+2)(k+3)}{3},\] which is the formula for \(n = k+1\).

By induction the formula holds for all \(n \ge 1\).

17 Induction: divisibility ★★★

Base case. \(4^1 - 1 = 3\), divisible by 3.

Inductive step. Assume \(4^k - 1 = 3m\) for some integer \(m\). Then \(4^{k+1} - 1 = 4\cdot 4^k - 1 = 4(4^k - 1) + 3 = 4(3m) + 3 = 3(4m + 1)\), which is divisible by 3.

By induction, \(3\) divides \(4^n - 1\) for all \(n \ge 1\).

18 Find the formula, then prove it ★★★

(a) \(a_2 = 7\), \(a_3 = 17\), \(a_4 = 37\). The formula gives \(5\cdot 2 - 3 = 7\), \(5\cdot 4 - 3 = 17\), \(5\cdot 8 - 3 = 37\), as expected.

(b) Base case. \(5\cdot 2^0 - 3 = 2 = a_1\).

Inductive step. If \(a_k = 5\cdot 2^{k-1} - 3\), then \(a_{k+1} = 2a_k + 3 = 10\cdot 2^{k-1} - 6 + 3 = 5\cdot 2^{k} - 3\), the formula for \(k+1\).

Hence \(a_n = 5\cdot 2^{n-1} - 3\) for all \(n \ge 1\).

19 The bouncing ball ★★★

The ball falls 10 ft once. Then, for each rebound, it goes up and comes down: the rebound heights are \(6, 3.6, 2.16, \ldots\), a geometric sequence with \(a_1 = 6\), \(r = 0.6\).

Sum of the rebound heights: \(\dfrac{6}{1 - 0.6} = 15\) ft. Each is traveled twice: \(2 \times 15 = 30\) ft.

Total: \(10 + 30 = 40\) ft, which is \(40 \times 0.3048 = 12.192 \approx 12.2\) m.

20 Finding a_1 and d ★★★

From \(a_9 - a_4 = 5d = 25\), we get \(d = 5\). Then \(a_1 = a_4 - 3d = 17 - 15 = 2\).

\(a_{30} = 2 + 29 \times 5 = 147\), so \(S_{30} = \dfrac{30(2 + 147)}{2} = 15 \times 149 = 2235\).

21 An unknown ratio ★★★

\(a_5 = 81 r^4 = 16\), so \(r^4 = \dfrac{16}{81}\) and \(r = \dfrac23\) or \(r = -\dfrac23\).

For \(r = \dfrac23\): terms 81, 54, 36, 24, 16, and \(S_5 = 211\).

For \(r = -\dfrac23\): terms 81, \(-54\), 36, \(-24\), 16, and \(S_5 = 55\).

Check with the formula for \(r = \dfrac23\): \(81\cdot\dfrac{1 - 32/243}{1/3} = 243 \cdot \dfrac{211}{243} = 211\).