If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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.
$\begin{cases} a(1)=\greenE 3 \\\\ a(n)=a(n-1)\maroonC{+2} \end{cases}$
Recall that this formula gives us the following two pieces of information:
• The first term is start color #0d923f, 3, end color #0d923f
• To get any term from its previous term, add start color #ed5fa6, 2, end color #ed5fa6. In other words, the common difference is start color #ed5fa6, 2, end color #ed5fa6.
Let's find an explicit formula for the sequence.
Remember that we can represent a sequence whose first term is start color #0d923f, A, end color #0d923f and common difference is start color #ed5fa6, B, end color #ed5fa6 with the standard explicit form start color #0d923f, A, end color #0d923f, plus, start color #ed5fa6, B, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis.
Therefore, an explicit formula of the sequence is a, left parenthesis, n, right parenthesis, equals, start color #0d923f, 3, end color #0d923f, start color #ed5fa6, plus, 2, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis.

1) Write an explicit formula for the sequence.
$\begin{cases} b(1)=-22 \\\\ b(n)=b(n-1)+7 \end{cases}$
b, left parenthesis, n, right parenthesis, equals

2) Write an explicit formula for the sequence.
$\begin{cases} c(1)=8 \\\\ c(n)=c(n-1)-13 \end{cases}$
c, left parenthesis, n, right parenthesis, equals

## 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.
d, left parenthesis, n, right parenthesis, equals, start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 16, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis
This formula is given in the standard explicit form start color #0d923f, A, end color #0d923f, plus, start color #ed5fa6, B, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis where start color #0d923f, A, end color #0d923f is the first term and that start color #ed5fa6, B, end color #ed5fa6 is the common difference. Therefore,
• the first term of the sequence is start color #0d923f, 5, end color #0d923f, and
• the common difference is start color #ed5fa6, 16, end color #ed5fa6.
Let's find a recursive formula for the sequence. Recall that the recursive formula gives us two pieces of information:
1. The first term left parenthesiswhich we know is start color #0d923f, 5, end color #0d923f, right parenthesis
2. The pattern rule to get any term from the term that comes before it left parenthesiswhich we know is "add start color #ed5fa6, 16, end color #ed5fa6"right parenthesis
Therefore, this is a recursive formula for the sequence.
$\begin{cases} d(1)=\greenE 5\\\\ d(n)=d(n-1)\maroonC{+16} \end{cases}$

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

We are given the following explicit formula of an arithmetic sequence.
e, left parenthesis, n, right parenthesis, equals, 10, plus, 2, n
Note that this formula is not given in the standard explicit form start color #0d923f, A, end color #0d923f, plus, start color #ed5fa6, B, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis.
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:
• e, left parenthesis, start color #11accd, 1, end color #11accd, right parenthesis, equals, 10, plus, 2, dot, start color #11accd, 1, end color #11accd, equals, 12
• e, left parenthesis, start color #11accd, 2, end color #11accd, right parenthesis, equals, 10, plus, 2, dot, start color #11accd, 2, end color #11accd, equals, 14
Now we can see that the first term is start color #0d923f, 12, end color #0d923f and the common difference is start color #ed5fa6, 2, end color #ed5fa6.
Therefore, this is a recursive formula for the sequence.
$\begin{cases} e(1)=\greenE{12}\\\\ e(n)=e(n-1)\maroonC{+2} \end{cases}$

3) The explicit formula of an arithmetic sequence is f, left parenthesis, n, right parenthesis, equals, 5, plus, 12, left parenthesis, n, minus, 1, right parenthesis.
Complete the missing values in the recursive formula of the sequence.
$\begin{cases} f(1)=A\\\\ f(n)=f(n-1)+B \end{cases}$
A, equals
B, equals

4) The explicit formula of an arithmetic sequence is g, left parenthesis, n, right parenthesis, equals, minus, 11, minus, 8, left parenthesis, n, minus, 1, right parenthesis.
Complete the missing values in the recursive formula of the sequence.
$\begin{cases} g(1)=A\\\\ g(n)=g(n-1)+B \end{cases}$
A, equals
B, equals

5) The explicit formula of an arithmetic sequence is h, left parenthesis, n, right parenthesis, equals, 1, plus, 4, n.
Complete the missing values in the recursive formula of the sequence.
$\begin{cases} h(1)=A\\\\ h(n)=h(n-1)+B \end{cases}$
A, equals
B, equals

6) The explicit formula of an arithmetic sequence is i, left parenthesis, n, right parenthesis, equals, 23, minus, 6, n.
Complete the missing values in the recursive formula of the sequence.
$\begin{cases} i(1)=A\\\\ i(n)=i(n-1)+B \end{cases}$
A, equals
B, equals

### Challenge problem

7*) Select all the formulas that correctly represent the arithmetic sequence 101, comma, 114, comma, 127, comma, point, point, point

## Want to join the conversation?

• which formula is more suitable to use recursive or explicit? • why will
​j(1)=114
​j(n)=j(n−1)+13
​​not work as an answer for 101 114 127? • What if there isn't a constant common difference;
say the number doubles every step?
We cannot plug in the value of B
Have a test coming up • 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? • Hi! How do you obtain an explicit formula from the recursive formula a_n+1=1/((a_n)+1) where a_1=1 • 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? • 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)
• 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
• 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) ?? • What about converting this recursive formula to an explicit one?
f(1) = 1f(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? • 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 