Write a recursive formula for the sequence.

Question

Write a recursive formula for the sequence. 

-9, 21, 51, 81, 111, ...

Write an explicit formula for the following sequence. 

-29, -25, -21, -17, -13, ...

Answer 1

Write a recursive formula for the sequence. 

-9, 21, 51, 81, 111, ...

So this is an arithmetic sequence, meaning that some number (positive or negative) is added to the number before it, and that number never changes. So look from -9 to 21. We went from smaller to bigger, so perhaps the pattern here is some positive number that is added to each term... -9 + 21 is 30 (to figure that out, you could do 21 - (-9) = 30). Also, 21 + 30 = 51, 51 + 30 = 81, etc. There is a pattern! 30 gets added to a term in the sequence to get the next term. Here's how we express this mathematically:

Let a_n be a term in the sequence (specifically, the n^th term), and let a_n+1 be the next term (one term after the previous term, hence the n + 1). Then:

a_n+1 = a_n + 30. <----- This is your recursive formula to get from one term to the term right after it.

How to use the formula?

The first term, or term n = 1, is -9. -9 + 30 = 21. This is the second term, or n+1 = 1 + 1, or n+1 = 2.

The second term, or term n = 2, is 21. 21 + 30 = 51. This is the third term, or n+1 = 2 + 1, or n+1 = 3.

And so forth.

There is a more general formula to get from the first term of the sequence to any term after it. This is why you might have an arithmetic sequence formula like this in your books:

a_n = a₁ + (n-1)d

a_n just refers to a number in the sequence, and a₁ is the first term of the sequence. The d refers to the number that keeps getting added (so here, d = 30), and n-1 is there to catapult us to some number in the sequence. For instance, if we want the 68th term of the sequence (which really means we're adding 30 to the first term of the sequence 67 times), here's what we'll do:

a_n = a₁ + (n-1)d

a₆₈ = (-9) + (68-1)(30) <--- since we want the 68th term of the sequence. -9 is the first number in the sequence, and 30 is the thing being added to each term

a₆₈ = -9 + (67)(30) <---- do this by hand or in the calculator correctly, remembering order of operations

a₆₈ = 2001 <----- better than adding 30 67 times, right? That's what the n-1 does for us.

So here's your most general recursive formula, and most likely the answer you need:

a_n = -9 + (n-1)(30)

I'll rearrange the terms a bit:

a_n = -9 + 30(n-1) <---- general recursive formula

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The second part of this problem looks exactly like this, and the pattern is clear. Your turn. :)

I hope this helps!

Answer 2

a₁ = - 9, a_n = a_n-1 + 30, n ≥ 2

b_n = - 29 + (n-1)·4 = 4n - 33, n ≥ 1

Answer 3

Observe that to get to the next term, you need to add 30.

So, a(1)=-9 (this means 1st term is -9)

and a(n)=a(n-1)+30 (this means the general term is the preceeding term plus 30)

For the second arithmetic sequence, a(n)=a(1)+(n-1)d (formula for the general term)

d=4 (to get to the next term, you need to add 4) (d is the common difference)

a(1)=-29 (this means 1st term is -29)

Plug in to get a(n)=-29+(n-1)(4)

Simplify to get a(n)=-29+4n-4

Simplify further to get a(n)=-33+4n

Answer 4

-9, 21, 51, 81, 111, ... is an Arithmetic Sequence with a common difference of 30

The Recursive Definition defines each value in terms of the previous value(s):

a₁ = -9

a_n = a_n-1 + 30

-29, -25, -21, -17, -13, ... is an Arithmetic Sequence with a common difference of 4.

The Explicit Definition defines each value in terms of its position in the sequence (that is, n):

a₁ = -29

a_{n =} 4n - 33 . . . . . . . . . . . . from: . a_n = a_n +d(n-1).. . . . . . . a_n = -29 + 4(n - 1)