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


Solution: 3, -6, 12, -24, 48, -96
 

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


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

3. c₁ = 2, c_n = 2c_n-1 -1 for n > 2.


Solution: 2, 3, 5, 9, 17, 33
 

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


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

2. a₁ = 5; a_n = 2·a_n-1


Solution: a_n = 5·2n-1
 

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

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

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 s_n = (a₁ + a_n)/ 2, the calculation is almost the same.
Alternatively, we have s_n = 4(1 + 2 + ... + 50) = 4·50·51/ 2 = 5100.

2. {1, 5, 9, ... , 93, 97, 101}

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 s_n = (a₁ + a_n)/ 2, the calculation is almost the same.
Alternatively, we have s_n = 1 + (1 + 4·1) + (1 + 4·2) + ... + (1 + 25·4).
Thus, s_n = 1·26 + 4(1 + 2 + ... + 25) = 26 + 4·25·26/ 2 = 1326.
 

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

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 b_n = b_n-1 +12.
5. Consider the sequence: {9, 6, 4, 8/ 3, ... } = {9·( (2/ 3)^k)}_k=0,1,2,...
1. Find the sum of the first 6 terms.

Solution: s = 9·(1 - (2/ 3)⁶)/(1 - (2/ 3)) = 9·(1 - 64/ 729)/(1 - (2/ 3)) = 665/ 27 = 24.63
 

2. Find the sum of the first n terms

Solution: s = 9·(1 - (2/ 3)ⁿ)/(1 - (2/ 3)) = 27·(1 - (2/ 3)ⁿ)
 

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

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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