1. Determine whether each of the following sequences is arithmetic or geometric. Find the initial term, the recursion and the explicit formula.
1. {a_n} = {2, 5, 8, 11, ... }

Solution: arithmetic. a₁ = 2. recursion: a_n = a_n-1 + 3. Explicit formula: a_n = 2 + 3(n-1).
2. {b_n} = {2, 6, 18, 54, ... }

Solution: geometric. a₁ = 2. recursion: a_n = 3·a_n-1. Explicit formula: a_n = 2·3^n-1.
3. {c_n} = {3, -1, -5, -9, ... }

Solution: arithmetic. c₁ = 3. recursion: a_n = a_n-1 - 4. Explicit formula: a_n = 3 - 4(n-1).
4. {f_n} = {a₂, a₅, a₈, a₁₁, ... }, where a_n is as given above.

Solution: arithmetic. f₁ = a₂ = 5. recursion: f_n = f_n-1 + 9. (Add 3+3+3 each time!)
Explicit formula: a_n = 2 + 3(n-1).
2. In each case, use the recursion to calculate the first six terms of each sequence:
1. a₁ = -3, a_n = a_n-1 + 5 for n > 2.

solution:

n	1	2	3	4	5	6
an	-3	2	7	12	17	22

2. b₁ = 6, b_n = b_n-1/ 2 for n > 2.

solution:

n	1	2	3	4	5	6
bn	6	3	3/2	3/4	3/8	3/16

3. p₁ = 1, p₂ = 2; p_n = p_n-1·p_n-2 -1 for n > 3.

solution:

n	1	2	3	4	5	6	7	8	9	10
pn	1	2	1	1	0	-1	-1	0	-1	-1

3. Given the recursion and initial term, find the explicit formula:
1. a₁ = 3; a_n = a_n-1 - 2

solution: Explicit formula: a_n = 2 - 3(n-1).
2. g₁ = 7; g_n = 2 g_n-1

solution: Explicit formula: a_n = 7·2^n-1.
4. Given the following arithmetic sequence, find the sum (i)by adding backward and forward,
   and (ii)by using the formula 1 + 2 + 3 + ... + (n-1) + n = n(n+1)/2:
1. {2, 5, 8, ... , 92, 95, 98}

Method 1:

	2	5	8	. . .	92	95	98
	98	95	92	. . .	8	5	2
sum	100	100	100	. . .	100	100	100

So, the sum in the table is 3300 = 33·100. (To find the number of columns, note that

2 + 3(n- 1) = 98. So, 3(n-1) = 96, or n- 1 = 32.)
The sum of the sequence is 3300/2 = 1650.
Method 2: s_n = 33· 2 + 3(1 + 2 + ... + 32) = 66 + 3·32·33/ 2 = 66 + 1584 = 1650.

2. {b_n} = {a₁, a₂, a₃,... , a_n} where a_n = a₁ + (k-1)d for k = 1, 2, ... , n.

Method 1:

	a1	a1 + d	a1 + 2d	. . .	a1 + (n-2)d	a1 + (n-1)d
	a1 + (n-1)d	a1 + (n-2)d	a1 + (n-3)d	. . .	a1 + d	a1
sum	2a1 + (n-1)d	2a1 + (n-1)d	2a1 + (n-1)d	. . .	2a1 + (n-1)d	2a1 + (n-1)d

There are ncolumns, so the sum of the table is 2na₁ + n(n-1)d.
So, the sum of the first nterms of the sequence is half of this, or 2na₁ + n(n-1)d/ 2.

5. Find the sum of the even integers starting with 500 and ending with 800.

Solution: Use the formula s_n = (a₁+a_n)/ 2. There are 151 terms (check this!!), so that the sum is 151(800+500)/ 2 = 98150.
6. Find the sum of the first 20 terms of each of the following sequences:
1. {g_n} = {3, 6, 12, ... }

Solution: sum = 3(1 - 2²⁰)/ (-1) = 3(2²⁰-1)
2. {h_n} = {1, 5, 25, 125, ... }

Solution: sum = 1·(1 - 5²⁰)/ (-4) = (5²⁰-1)/ 4
7. Find the sum of the first nterms of each of the following sequences:
1. {g_n} = {1, 2/3, 4/9, 8/27, ... }

Solution: sum = 1·(1 - (2/3)ⁿ)/(1 - (2/3)) = (1 - (2/3)ⁿ)/(1/3), so that:
Solution: sum = (1 - (2/3)ⁿ)·(3/1) = 3·(1 - (2/3)ⁿ)
2. {h_n} = {1, .9, .81, .729, ... }

Solution: sum = 1·(1 - .9ⁿ)/ (1 - .9) = (1 - .9ⁿ)/ .1 = 10(1 - .9ⁿ)
8. Solution:As ntends to infinity, (2/3)ⁿ and 9ⁿ both tend to 0. So the first sum tends to 3, and the second sum tends to 10.

Back to the review questions.

For questions: roberts@math.umn.edu