### Fall 1999, test #1 solutions

1. In each case, use the recursion to find the first 6 terms of the sequence:.
1. g1 = 3, gn = -2gn-1 for n > 2

2. Solution: 3, -6, 12, -24, 48, -96

3. a1 = 3, an = an-1 + 2/ 3 for n > 2

4. Solution: 1, 5/ 3, 7/ 3, 3, 11/ 3, 13/ 3

5. c1 = 2, cn = 2cn-1 -1 for n > 2.

6. Solution: 2, 3, 5, 9, 17, 33

2. Given the following recursions, find the explicit formulas:
1. a1 = -3; an = an-1 + 2

2. Solution: an = -3 + 2(n-1) or an = 2n-5

3. a1 = 5; an = 2·an-1

4. Solution: an = 5·2n-1

3. Find the sums of the following arithmetic sequences (i)by adding backward and forward,

4. and (ii)by using one of the formulas (of your own choice):
1. {4, 8, 12, ... , 192, 196, 200}

2. Method 1:
 4 8 12 . . . 192 196 200 200 196 192 . . . 12 8 4 Sum 204 204 204 . . . 204 204 204

There are 50 terms, so the sum is (50·204)/ 2 = 5100.

Method 2: If we use the formula sn = (a1 + an)/ 2, the calculation is almost the same.
Alternatively, we have sn = 4(1 + 2 + ... + 50) = 4·50·51/ 2 = 5100.

3. {1, 5, 9, ... , 93, 97, 101}

4. Method 1:
 1 5 9 . . . 93 97 101 101 97 93 . . . 9 5 1 Sum 102 102 102 . . . 102 102 102

There are 26 terms, so the sum is (26·102)/ 2 = 1326.

Method 2: If we use the formula sn = (a1 + an)/ 2, the calculation is almost the same.
Alternatively, we have sn = 1 + (1 + 4·1) + (1 + 4·2) + ... + (1 + 25·4).
Thus, sn = 1·26 + 4(1 + 2 + ... + 25) = 26 + 4·25·26/ 2 = 1326.

5. Let the arithmetic sequence {a1 ,a2, a3, ... } have initial term a1 = 5 and recursion an = an-1 +4,

6. and let {b1 ,b2, b3, ... } = {a10 ,a13, a16, ... }. Find the initial term and recursion for the sequence
{b1 ,b2, b3, ... }.
Solution: The initial term is b1 = a10 = 41. The common difference is 4 + 4 + 4 = 12, so that the recursion is bn = bn-1 +12.
7. Consider the sequence: {9, 6, 4, 8/ 3, ... } = {( (2/ 3)k)}k=0,1,2,...
1. Find the sum of the first 6 terms.
2. Solution: s = 9·(1 - (2/ 3)6)/(1 - (2/ 3)) = 9·(1 - 64/ 729)/(1 - (2/ 3)) = 665/ 27 = 24.63

3. Find the sum of the first n terms
4. Solution: s = 9·(1 - (2/ 3)n)/(1 - (2/ 3)) = 27·(1 - (2/ 3)n)

5. Explain what happens to the value of the sum in part (b) as n tends to infinity,

6. and find the infinite sum 9 + 6 + 4 + 8/ 3 +

Solution: As n tends to infinity, (2/ 3)n tends to 0. So, the value of the sum of the first nterms tends to 27. So, the infinite sum is 27.

For questions roberts@math.umn.edu

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