 ### Solutions to Chapter 1 review questions 1. Determine whether each of the following sequences is arithmetic or geometric. Find the initial term, the recursion and the explicit formula.
1. {an} = {2, 5, 8, 11, ... }

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

4. Solution: geometric. a1 = 2. recursion: an = 3·an-1. Explicit formula: an = 2·3n-1.
5. {cn} = {3, -1, -5, -9, ... }

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

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

2. solution:
 n 1 2 3 4 5 6 an -3 2 7 12 17 22
3. b1 = 6, bn = bn-1/ 2 for n > 2.

4. solution:
 n 1 2 3 4 5 6 bn 6 3 3/2 3/4 3/8 3/16
5. p1 = 1, p2 = 2; pn = pn-1·pn-2 -1 for n > 3.

6. 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; an = an-1 - 2

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

4. solution: Explicit formula: an = 7·2n-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}

2. 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: sn = 33· 2 + 3(1 + 2 + ... + 32) = 66 + 3·32·33/ 2 = 66 + 1584 = 1650.

3. {bn} = {a1, a2, a3,... , an} where an = a1 + (k-1)d for k = 1, 2, ... , n.

4. 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.

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

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

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

2. 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)
3. {hn} = {1, .9, .81, .729, ... }

4. Solution: sum = 1·(1 - .9n)/ (1 - .9) = (1 - .9n)/ .1 = 10(1 - .9n)
9. 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