which formula is more suitable to use recursive or explicit?

I assume that the recursive is useful when we can't express a sequence in the explicit.

why will j(1)=114 j(n)=j(n−1)+13 not work as an answer for 101 114 127?

Because the first term is 101 and not 114

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

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

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?

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.

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) ??

The explicit formula in standard form is a(n)=a(1)+b(n-1).

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?

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

A10 how would you solve this, the 10 is suppose to be a lower n

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

Main content

Converting recursive & explicit forms of arithmetic sequences

Google Classroom

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.

Check your understanding

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

[I need help!]

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

[I need help!]

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:

The first term left parenthesiswhich we know is start color #0d923f, 5, end color #0d923f, right parenthesis
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}

Check your understanding

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 Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text	B, equals Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

[I need help!]

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 Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text	B, equals Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

[I need help!]

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 Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text	B, equals Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

[I need help!]

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 Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text	B, equals Your answer should be an integer, like 6 a simplified proper fraction, like 3, slash, 5 a simplified improper fraction, like 7, slash, 4 a mixed number, like 1, space, 3, slash, 4 an exact decimal, like 0, point, 75 a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

[I need help!]

Challenge problem

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

(Choice A)
$\begin{cases} j(1)=13\\\\ j(n)=j(n-1)+101 \end{cases}$
(Choice B)
$\begin{cases} j(1)=101\\\\ j(n)=j(n-1)+13 \end{cases}$
(Choice C)
$\begin{cases} j(1)=114\\\\ j(n)=j(n-1)+13 \end{cases}$
(Choice D)
j, left parenthesis, n, right parenthesis, equals, 101, plus, 13, n
(Choice E)
j, left parenthesis, n, right parenthesis, equals, 88, plus, 13, n
(Choice F)
j, left parenthesis, n, right parenthesis, equals, 114, plus, 13, n
(Choice G)
j, left parenthesis, n, right parenthesis, equals, 101, plus, 13, left parenthesis, n, minus, 1, right parenthesis

[I need help!]

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anirudhcr7
Posted 8 years ago. Direct link to anirudhcr7's post “which formula is more sui...”
which formula is more suitable to use recursive or explicit?
Button navigates to signup pageComment on anirudhcr7's post “which formula is more sui...”
(32 votes)
Answer
- Jenna Sep
  Posted 8 years ago. Direct link to Jenna Sep's post “I assume that the recursi...”
  I assume that the recursive is useful when we can't express a sequence in the explicit.
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  (13 votes)
anaseg18
Posted 7 years ago. Direct link to anaseg18's post “why will j(1)=114 j(n)=...”
why will
j(1)=114
j(n)=j(n−1)+13
not work as an answer for 101 114 127?
Button navigates to signup pageComment on anaseg18's post “why will j(1)=114 j(n)=...”
(9 votes)
Answer
- deepesh k
  Posted 7 years ago. Direct link to deepesh k's post “Because the first term is...”
  Because the first term is 101 and not 114
  Comment on deepesh k's post “Because the first term is...”
  (9 votes)
Narun Singh
Posted 5 years ago. Direct link to Narun Singh's post “What if there isn't a con...”
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
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(5 votes)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “Arithmetic sequences use ...”
  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
  Button navigates to signup page
  (7 votes)
callie
Posted 2 months ago. Direct link to callie's post “In question 7, when simpl...”
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?
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(4 votes)
Answer
- Tanner P
  Posted 2 months ago. Direct link to Tanner P's post “In the 13(n - 1), the 13 ...”
  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.
  Button navigates to signup page
  (3 votes)
Hope Wilson
Posted 4 years ago. Direct link to Hope Wilson's post “Hi! How do you obtain an ...”
Hi! How do you obtain an explicit formula from the recursive formula a_n+1=1/((a_n)+1) where a_1=1
Button navigates to signup pageComment on Hope Wilson's post “Hi! How do you obtain an ...”
(3 votes)
Answer
ShelbyMFleming
Posted 5 years ago. Direct link to ShelbyMFleming's post “What if it doesn't provid...”
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!!
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(2 votes)
Answer
- Selina Sun
  Posted 9 months ago. Direct link to Selina Sun's post “I am providing this answe...”
  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!
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  (1 vote)
denisejwagner
Posted 3 years ago. Direct link to denisejwagner's post “A10 how would you solve t...”
A10 how would you solve this, the 10 is suppose to be a lower n
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(1 vote)
Answer
- Jerry Nilsson
  Posted 3 years ago. Direct link to Jerry Nilsson's post “An arithmetic sequence is...”
  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
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  (4 votes)
Giannis Theodorakis
Posted 5 years ago. Direct link to Giannis Theodorakis's post “Is it right if we say tha...”
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) ??
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(2 votes)
Answer
- Nikolay
  Posted 3 years ago. Direct link to Nikolay's post “The explicit formula in s...”
  The explicit formula in standard form is a(n)=a(1)+b(n-1).
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  (2 votes)
Jasur Yusupov
Posted 3 years ago. Direct link to Jasur Yusupov's post “What about converting thi...”
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?
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(2 votes)
Answer
- Renz
  Posted a month ago. Direct link to Renz's post “Let's first try to expand...”
  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
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  (2 votes)
Red Arrow ✝
Posted 4 years ago. Direct link to Red Arrow ✝'s post “I had this question: ...”
I had this question:

f(1)= −71
f(n) = f(n−1)⋅4.2
Find an explicit formula for f(n)

And I answered f(n) = -71⋅4.2(n-1), but it said the answer was
f(n) = -71⋅4.2^(n-1). I haven't done recursive/explicit forms before (I guess I just skipped them in Algebra 1) so could someone shed some light here?
Button navigates to signup pageComment on Red Arrow ✝'s post “I had this question: ...”
(2 votes)
Answer

Algebra 1

Course: Algebra 1 > Unit 9

Converting from a recursive formula to an explicit formula

Check your understanding

Converting from an explicit formula to a recursive formula

Example 1: Formula is given in standard form

Example 2: Formula is given in simplified form

Check your understanding

Challenge problem

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