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### Course: Get ready for Precalculus > Unit 6

Lesson 2: Constructing arithmetic sequences- Recursive formulas for arithmetic sequences
- Recursive formulas for arithmetic sequences
- Recursive formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Arithmetic sequence problem
- Converting recursive & explicit forms of arithmetic sequences
- Converting recursive & explicit forms of arithmetic sequences
- Converting recursive & explicit forms of arithmetic sequences
- Arithmetic sequences review

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# Converting recursive & explicit forms of arithmetic sequences

Learn how to convert between recursive and explicit formulas of arithmetic sequences.

Before taking this lesson, make sure you know how to find recursive formulas and explicit formulas of arithmetic sequences.

## Converting from a recursive formula to an explicit formula

An arithmetic sequence has the following recursive formula.

Recall that this formula gives us the following two pieces of information:

- The first term is
${3}$ - To get any term from its previous term,
**add**In other words, the common difference is .${2}$ .${2}$

Let's find an explicit formula for the sequence.

Remember that we can represent a sequence whose first term is ${A}$ and common difference is ${B}$ with the standard explicit form ${A}+{B}(n-1)$ .

Therefore, an explicit formula of the sequence is $a(n)={3}{+2}(n-1)$ .

### Check your understanding

## Converting from an explicit formula to a recursive formula

### Example 1: Formula is given in standard form

We are given the following explicit formula of an arithmetic sequence.

This formula is given in the standard explicit form ${A}+{B}(n-1)$ where ${A}$ is the first term and that ${B}$ is the common difference. Therefore,

- the first term of the sequence is
, and${5}$ - the common difference is
.${16}$

Let's find a recursive formula for the sequence. Recall that the recursive formula gives us two pieces of information:

- The first term
which we know is$($ ${5})$ - The pattern rule to get any term from the term that comes before it
which we know is "add$($ "${16}$ $)$

Therefore, this is a recursive formula for the sequence.

### Example 2: Formula is given in simplified form

We are given the following explicit formula of an arithmetic sequence.

Note that this formula is ${A}+{B}(n-1)$ .

*given in the standard explicit form***not**For this reason, we can't simply use the structure of the formula to find the first term and the common difference. Instead, we can find the first two terms:

Now we can see that the first term is ${12}$ and the common difference is ${2}$ .

Therefore, this is a recursive formula for the sequence.

### Check your understanding

### Challenge problem

## Want to join the conversation?

- which formula is more suitable to use recursive or explicit?(33 votes)
- I assume that the recursive is useful when we can't express a sequence in the explicit.(17 votes)

- why will

j(1)=114

j(n)=j(n−1)+13

not work as an answer for 101 114 127?(11 votes)- Because the first term is 101 and not 114(14 votes)

- What if there isn't a constant common difference;

say the number doubles every step?

We cannot plug in the value of B

Please Help!!

Have a test coming up(5 votes)- Arithmetic sequences use addition or subtraction to get the next term in the sequence. It sounds like you have a geometric sequence which uses multiplication or division to get to the next item in the sequence. Geometric sequences are covered at this link: https://www.khanacademy.org/math/algebra/sequences/introduction-to-geometric-sequences/v/geometric-sequences-introduction(8 votes)

- In question 7, when simplifying the formula, how does
**j(n)=101+13(n−1) become**?

101+13n−13

Where did the -13 come from?(5 votes)- In the 13(n - 1), the 13 was distributed to the n and the -1.

13(n-1) = 13n - 13

Remember that when you use the distributive property, you have to consider every term (and its sign) inside the parentheses.(6 votes)

- Is it right if we say that when we are given a formula in simlified form like the one in exapmle two ( f(n)=a+bn ), the explicit form of the sequence is h(n)=(a+b)+b(n-1) because:

a+bn= a+b-b+bn = (a+b)+b(n-1) ??(3 votes)- The explicit formula in standard form is a(n)=a(1)+b(n-1).(2 votes)

- What about converting this recursive formula to an explicit one?
`f(1) = 1`

f(n) = f(n-1) + 2n

I think that this solution is**incorrect**:`1+2n(n-1)`

So, what formula should we use when there are`kn`

or`n/k`

, instead of simple`k`

in the common difference part?(3 votes)- Let's first try to expand the recursion formula by plugging in an actual number for n, say n = 5

f(n) = f(n-1) + 2n

f(5) = f(4) + 10 = 29

f(4) = f(3) + 8 = 19

f(3) = f(2) + 6 = 11

f(2) = f(1) + 4 = 5

f(1) = 1, given

As we can see, the equations above do not exactly describe an arithmetic sequence.

But we can observe something interesting about their differences (ie. 29 minus 19, 19 minus 11, etc. ). As n increases the

difference between the terms is incremented by 2.

This is interesting because now we have a clue about what the explicit formula could look like.

The formula may be in a form similar to arithmetic progression:

a(n) = a(1) + d(n - 1)

where a(1) is the initial term, d is the common difference and a(n) is the n-th term of the sequence

To obtain the explicit formula we must find a way to generalize the equations above into a single equation.

We may know that

f(n) = f(n-1) + 2n = f(n-1) + 2(n-0)

f(n-1) = f(n-1-1) + 2(n-1-0) = f(n-2) + 2(n-1)

f(n) = (f(n-2) + 2(n-1)) + 2(n-0)

f(n-2) = f(n-3) + 2(n-2)

f(n) = (f(n-3) + 2(n-2)) + 2(n-1) + 2(n-0)

f(n-3) = f(n-4) + 2(n-3)

f(n) = (f(n-4) + 2(n-3)) + 2(n-2) + 2(n-1) + 2(n-0)

let's check for n = 5

f(5) = (f(5-4) + 2(5-3)) + 2(5-2) + 2(5-1) + 2(5-0)

f(5) = (f(1) + 2(2)) + 2(3) + 2(4) + 2(5)

f(5) = 29

by induction we find that

f(n) = f(n-m) + 2(n-(m-1)) + 2(n-(m-2)) + 2(n-(m-3)) +...+ 2(n)

where m = n - 1

f(n) = f(n-(n-1)) + 2(n-(n-1-1)) + 2(n-(n-1-2)) + 2(n-(n-1-3)) +...+ 2(n)

f(n) = f(1) + 2(2) + 2(3) + 2(4) +...+ 2(n)

f(n) = 1 + 2(k=2∑n k)

we can evaluate the last equation to a more workable formula that uses sigma notation for arithmetic series.

f(n) = 1 + 2((k=0∑n k) - 1)

f(n) = 1 + 2((n(n+1))/2) - 2

f(n) = n(n+1) - 1

f(n) = n² + n - 1, for all positive n integer

let's check for n = 5

f(5) = 5² + 5 - 1

f(5) = 29(2 votes)

- A10 how would you solve this, the 10 is suppose to be a lower n(1 vote)
- An arithmetic sequence is defined recursively as

𝑎(𝑛 + 1) = 𝑎(𝑛) + 𝑑, for 𝑛 ≥ 1

From that we can derive the explicit formula.

𝑎(2) = 𝑎(1) + 𝑑

𝑎(3) = 𝑎(2) + 𝑑 = 𝑎(1) + 2𝑑

𝑎(4) = 𝑎(3) + 𝑑 = 𝑎(1) + 3𝑑

𝑎(5) = 𝑎(4) + 𝑑 = 𝑎(1) + 4𝑑

⋮

𝑎(𝑛) = 𝑎(1) + (𝑛 − 1)𝑑

So, as long as we know the first term, 𝑎(1), and the common difference, 𝑑, we can use the explicit formula to find any term of the sequence.

𝑛 = 10 ⇒ 𝑎(10) = 𝑎(1) + (10 − 1)𝑑 = 𝑎(1) + 9𝑑

– – –

Example:

Find the 10th term of the arithmetic sequence {−8, −5, −2, ...}

The first term is

𝑎(1) = −8

The common difference is

𝑑 = −5 − (−8) = 3

Thereby, the 10th term is

𝑎(10) = −8 + (10 − 1) ∙ 3 = 19(5 votes)

- Hi! How do you obtain an explicit formula from the recursive formula a_n+1=1/((a_n)+1) where a_1=1(3 votes)
- What if it doesn't provide a(1)? My question says "What is the recursive formula for the sequence defined by the explicit equation a(n)=15-3n?

Please help!!(2 votes)- I am providing this answer for anyone else who might need it. so if you need to find the a(1) and the common difference, you must use the standard explicit formula such a(n)=a(1)+d(common difference)*(n-1). So when you have equivalent formulas, you need to reform it back to the standard formula to be able to find the a(1) and d (common difference). In your case, I will reformat a(n)=15-3n as follows: a(n)=15-3(n-1)-3 ==> a(n)=12-3(n-1). As you can see, this final conversion is the standard explicit equation. so a(1)=12 and common difference = -3. Hope this helps someone!(1 vote)

- i(n)=23-6n ??

T^T idk wat to do...(2 votes)- i(n)=23-6n => i(n)= ? -6(n-1) => i(n)=17-6(n-1) So,A=17,B=-6(1 vote)