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

Course: Algebra 1 (OPS pilot — textbook aligned) > Unit 4

Lesson 6: Arithmetic sequences | 4-7

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.

${\begin{cases} a (1) = 3 \\ a (n) = a (n - 1) + 2 \end{cases}$ ‍

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 $2$ ‍. In other words, the common difference is $2$ ‍.

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

1) Write an explicit formula for the sequence.

{\begin{cases} b (1) = - 22 \\ b (n) = b (n - 1) + 7 \end{cases}

b (n) =

2) Write an explicit formula for the sequence.

{\begin{cases} c (1) = 8 \\ c (n) = c (n - 1) - 13 \end{cases}

c (n) =

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.

$d (n) = 5 + 16 (n - 1)$ ‍

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 $5$ ‍, and
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.

{\begin{cases} d (1) = 5 \\ d (n) = d (n - 1) + 16 \end{cases}

Example 2: Formula is given in simplified form

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

$e (n) = 10 + 2 n$ ‍

Note that this formula is not given in the standard explicit form

A + B (n - 1)

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:

$e (1) = 10 + 2 \cdot 1 = 12$ ‍
$e (2) = 10 + 2 \cdot 2 = 14$ ‍

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.

{\begin{cases} e (1) = 12 \\ e (n) = e (n - 1) + 2 \end{cases}

Check your understanding

3) The explicit formula of an arithmetic sequence is

f (n) = 5 + 12 (n - 1)

Complete the missing values in the recursive formula of the sequence.

{\begin{cases} f (1) = A \\ f (n) = f (n - 1) + B \end{cases}


$A =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍	$B =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍

4) The explicit formula of an arithmetic sequence is

g (n) = - 11 - 8 (n - 1)

Complete the missing values in the recursive formula of the sequence.

{\begin{cases} g (1) = A \\ g (n) = g (n - 1) + B \end{cases}


$A =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍	$B =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍

5) The explicit formula of an arithmetic sequence is

h (n) = 1 + 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}


$A =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍	$B =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍

6) The explicit formula of an arithmetic sequence is

i (n) = 23 - 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}


$A =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍	$B =$ ‍ Your answer should be an integer, like $6$ ‍ a simplified proper fraction, like $3 / 5$ ‍ a simplified improper fraction, like $7 / 4$ ‍ a mixed number, like $1 3 / 4$ ‍ an exact decimal, like $0.75$ ‍ a multiple of pi, like $12 pi$ ‍ or $2 / 3 pi$ ‍

Challenge problem

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

The first term of the sequence is

101

, and the common difference is

13

Therefore, this is the correct recursive formula for the sequence.

${\begin{cases} j (1) = 101 \\ j (n) = j (n - 1) + 13 \end{cases}$ ‍

The other recursive formulas have the wrong first term.

This is the standard form explicit formula for the sequence.

$j (n) = 101 + 13 (n - 1)$ ‍

We can simplify it to find another correct formula.

$\begin{aligned} j (n) & = 101 + 13 (n - 1) \\ = 101 + 13 n - 13 \\ = 88 + 13 n \end{aligned}$ ‍

The other explicit formulas are not equivalent to the above formula, and therefore are incorrect.

In conclusion, these are the correct formulas:

${\begin{cases} j (1) = 101 \\ j (n) = j (n - 1) + 13 \end{cases}$ ‍
$j (n) = 101 + 13 (n - 1)$ ‍
$j (n) = 88 + 13 n$ ‍

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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...”
(33 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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  (17 votes)
anaseg18
Posted 8 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)=...”
(11 votes)
Answer
- deepesh k
  Posted 8 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...”
  (14 votes)
Narun Singh
Posted 6 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 6 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
  (8 votes)
callie
Posted 9 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?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Tanner P
  Posted 9 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
  (6 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) ??
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Nikolay
  Posted 4 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).
  Button navigates to signup page
  (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?
Button navigates to signup pageComment on Jasur Yusupov's post “What about converting thi...”
(3 votes)
Answer
- Renz
  Posted 8 months 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
  Button navigates to signup page
  (2 votes)
denisejwagner
Posted 4 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
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Jerry Nilsson
  Posted 4 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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  (5 votes)
Hope Wilson
Posted 5 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 6 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 a year 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)
vivian
Posted 5 months ago. Direct link to vivian's post “i(n)=23-6n ?? T^T idk ...”
i(n)=23-6n ??

T^T idk wat to do...
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(2 votes)
Answer
- Ava
  Posted 4 months ago. Direct link to Ava's post “i(n)=23-6n => i(n)= ? -6(...”
  i(n)=23-6n => i(n)= ? -6(n-1) => i(n)=17-6(n-1) So,A=17,B=-6
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  (1 vote)