Difficulty understanding what the (n-1) part refers to in a basic arithmetic sequence
$begingroup$
I am 28 and learned no maths in school. I'm going back through and learning on my own time. I am up to arithmetic sequences. The general formula (according to Khan Academy) for a recursive arithmetic sequence seems to be:
a(1)=3
a(n)=a(n−1)+2
I have a basic understanding of the concept of a sequence, i.e. that it is describing a relationship between numbers that involves either repeated addition or subtraction (it may involve multiplication and division as well but I'm not sure as I'm not up to that).
I cannot understand what the (n-1) is. I get that the a(1) is the first term and that the +2 at the end is what gets added to the term each time. But what is the n-1. I know it refers to something along the lines of finding the term before the term, I think?
Cheers for any explanations or examples.
sequences-and-series
asked Jan 15 at 23:44
lachylachy
111
$endgroup$
add a comment |
$begingroup$
I am 28 and learned no maths in school. I'm going back through and learning on my own time. I am up to arithmetic sequences. The general formula (according to Khan Academy) for a recursive arithmetic sequence seems to be:
a(1)=3
a(n)=a(n−1)+2
I have a basic understanding of the concept of a sequence, i.e. that it is describing a relationship between numbers that involves either repeated addition or subtraction (it may involve multiplication and division as well but I'm not sure as I'm not up to that).
I cannot understand what the (n-1) is. I get that the a(1) is the first term and that the +2 at the end is what gets added to the term each time. But what is the n-1. I know it refers to something along the lines of finding the term before the term, I think?
Cheers for any explanations or examples.
sequences-and-series
asked Jan 15 at 23:44
lachylachy
111
$endgroup$
$begingroup$
Your age and background are unimportant.
$endgroup$
– John Douma
Jan 16 at 0:31
$begingroup$
Just like we use $x$ to refer to non-specific number when we don't know which one we use $a_n$ to refer to one of the terms in some position but we don't know which position. We call it the $n$th position. So $a_{n-1}$ is the term in the position immediately before it. And $a_{n+1}$ is the term in the position immediately after it.
$endgroup$
– fleablood
Jan 16 at 0:54
add a comment |
$begingroup$
I am 28 and learned no maths in school. I'm going back through and learning on my own time. I am up to arithmetic sequences. The general formula (according to Khan Academy) for a recursive arithmetic sequence seems to be:
a(1)=3
a(n)=a(n−1)+2
I have a basic understanding of the concept of a sequence, i.e. that it is describing a relationship between numbers that involves either repeated addition or subtraction (it may involve multiplication and division as well but I'm not sure as I'm not up to that).
I cannot understand what the (n-1) is. I get that the a(1) is the first term and that the +2 at the end is what gets added to the term each time. But what is the n-1. I know it refers to something along the lines of finding the term before the term, I think?
Cheers for any explanations or examples.
sequences-and-series
asked Jan 15 at 23:44
lachylachy
111
$endgroup$
I am 28 and learned no maths in school. I'm going back through and learning on my own time. I am up to arithmetic sequences. The general formula (according to Khan Academy) for a recursive arithmetic sequence seems to be:
a(1)=3
a(n)=a(n−1)+2
I have a basic understanding of the concept of a sequence, i.e. that it is describing a relationship between numbers that involves either repeated addition or subtraction (it may involve multiplication and division as well but I'm not sure as I'm not up to that).
I cannot understand what the (n-1) is. I get that the a(1) is the first term and that the +2 at the end is what gets added to the term each time. But what is the n-1. I know it refers to something along the lines of finding the term before the term, I think?
Cheers for any explanations or examples.
sequences-and-series
sequences-and-series
asked Jan 15 at 23:44
lachylachy
111
asked Jan 15 at 23:44
lachylachy
111
asked Jan 15 at 23:44
lachylachy
111
asked Jan 15 at 23:44
lachylachy
111
asked Jan 15 at 23:44
lachylachy
111
111
$begingroup$
Your age and background are unimportant.
$endgroup$
– John Douma
Jan 16 at 0:31
$begingroup$
Just like we use $x$ to refer to non-specific number when we don't know which one we use $a_n$ to refer to one of the terms in some position but we don't know which position. We call it the $n$th position. So $a_{n-1}$ is the term in the position immediately before it. And $a_{n+1}$ is the term in the position immediately after it.
$endgroup$
– fleablood
Jan 16 at 0:54
add a comment |
$begingroup$
Your age and background are unimportant.
$endgroup$
– John Douma
Jan 16 at 0:31
$begingroup$
Just like we use $x$ to refer to non-specific number when we don't know which one we use $a_n$ to refer to one of the terms in some position but we don't know which position. We call it the $n$th position. So $a_{n-1}$ is the term in the position immediately before it. And $a_{n+1}$ is the term in the position immediately after it.
$endgroup$
– fleablood
Jan 16 at 0:54
$begingroup$
Your age and background are unimportant.
$endgroup$
– John Douma
Jan 16 at 0:31
$begingroup$
Your age and background are unimportant.
$endgroup$
– John Douma
Jan 16 at 0:31
$begingroup$
Just like we use $x$ to refer to non-specific number when we don't know which one we use $a_n$ to refer to one of the terms in some position but we don't know which position. We call it the $n$th position. So $a_{n-1}$ is the term in the position immediately before it. And $a_{n+1}$ is the term in the position immediately after it.
$endgroup$
– fleablood
Jan 16 at 0:54
$begingroup$
Just like we use $x$ to refer to non-specific number when we don't know which one we use $a_n$ to refer to one of the terms in some position but we don't know which position. We call it the $n$th position. So $a_{n-1}$ is the term in the position immediately before it. And $a_{n+1}$ is the term in the position immediately after it.
$endgroup$
– fleablood
Jan 16 at 0:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Recursive functions define values in terms of previous values.
In your case, the definition of $a$ says that the first value is $3$ and we get the $n$th value by adding $2$ to the previous value.
So
$a(2)=a(1)+2=3+2=5$,
$a(3)=a(2)+2=5+2=7$, etc.
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
$endgroup$
add a comment |
$begingroup$
Your sequence will be a bunch of terms: $a_1, a_2, a_3, a_4,...... $ and we can refer to each term by its index. The fifth term is $a_5$. The $39$th term is $a_{39}$ and so on.
If we pick some arbitrary term but want to indicate it might be any term we can call it be an arbitrary index. $a_n$ is the $n$-th term. It could be the $512$th term, $a_{512}$ or it could by the $7,487$th term, $a_{7,487}$. We don't know and don't care. We are refering to a non-specific term.
We are refering to a term when we don't know what position it is in. We say "let's call the position it is in: $n$".
Now, let's suppose I said "After picking that term look at the previous term" or "look at the next term". How can we refer to that. We if our term was the $a_{512}$ then the previous term would be .... $a_{511}$. And if our term was $a_{7,487}$ the next term would be .... $a_{7,488}$.
But how do we refer to it in general. If our term was $a_n$ what is the term immediately before it. Well, if this was the $n$th term, the one write before it is the $n-1$th term, $a_{n-1}$.
So in this case:
$a_n = a_{n-1} + 2$
means. "Pick any term. It is equal to the term before it plus two".
So if the previous term $a_{n-1}$ was equal to $517$ then this term, $a_n$ will be equal to $519$.
So $a_1 = 3$ because we were told so.
And $a_2 = $ the previous term plus $2 = 3+2 = 5$.
And $a_3 = a_2 + 2 = 5 + 2 = 7$ and so on...
$a_4 = a_3 + 2 = 7 + 2= 9$ .....
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
$endgroup$
$begingroup$
Thanks very much for these helpful responses. This gives me a far better understanding of where the n-1 is coming from. I appreciate the time you all took to post. Cheers
$endgroup$
– lachy
Jan 17 at 0:26
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
Recursive functions define values in terms of previous values.
In your case, the definition of $a$ says that the first value is $3$ and we get the $n$th value by adding $2$ to the previous value.
So
$a(2)=a(1)+2=3+2=5$,
$a(3)=a(2)+2=5+2=7$, etc.
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
$endgroup$
add a comment |
$begingroup$
Recursive functions define values in terms of previous values.
In your case, the definition of $a$ says that the first value is $3$ and we get the $n$th value by adding $2$ to the previous value.
So
$a(2)=a(1)+2=3+2=5$,
$a(3)=a(2)+2=5+2=7$, etc.
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
$endgroup$
add a comment |
$begingroup$
Recursive functions define values in terms of previous values.
In your case, the definition of $a$ says that the first value is $3$ and we get the $n$th value by adding $2$ to the previous value.
So
$a(2)=a(1)+2=3+2=5$,
$a(3)=a(2)+2=5+2=7$, etc.
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
$endgroup$
Recursive functions define values in terms of previous values.
In your case, the definition of $a$ says that the first value is $3$ and we get the $n$th value by adding $2$ to the previous value.
So
$a(2)=a(1)+2=3+2=5$,
$a(3)=a(2)+2=5+2=7$, etc.
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
answered Jan 16 at 0:25
John DoumaJohn Douma
5,76511520
5,76511520
add a comment |
add a comment |
$begingroup$
Your sequence will be a bunch of terms: $a_1, a_2, a_3, a_4,...... $ and we can refer to each term by its index. The fifth term is $a_5$. The $39$th term is $a_{39}$ and so on.
If we pick some arbitrary term but want to indicate it might be any term we can call it be an arbitrary index. $a_n$ is the $n$-th term. It could be the $512$th term, $a_{512}$ or it could by the $7,487$th term, $a_{7,487}$. We don't know and don't care. We are refering to a non-specific term.
We are refering to a term when we don't know what position it is in. We say "let's call the position it is in: $n$".
Now, let's suppose I said "After picking that term look at the previous term" or "look at the next term". How can we refer to that. We if our term was the $a_{512}$ then the previous term would be .... $a_{511}$. And if our term was $a_{7,487}$ the next term would be .... $a_{7,488}$.
But how do we refer to it in general. If our term was $a_n$ what is the term immediately before it. Well, if this was the $n$th term, the one write before it is the $n-1$th term, $a_{n-1}$.
So in this case:
$a_n = a_{n-1} + 2$
means. "Pick any term. It is equal to the term before it plus two".
So if the previous term $a_{n-1}$ was equal to $517$ then this term, $a_n$ will be equal to $519$.
So $a_1 = 3$ because we were told so.
And $a_2 = $ the previous term plus $2 = 3+2 = 5$.
And $a_3 = a_2 + 2 = 5 + 2 = 7$ and so on...
$a_4 = a_3 + 2 = 7 + 2= 9$ .....
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
$endgroup$
$begingroup$
Thanks very much for these helpful responses. This gives me a far better understanding of where the n-1 is coming from. I appreciate the time you all took to post. Cheers
$endgroup$
– lachy
Jan 17 at 0:26
add a comment |
$begingroup$
Your sequence will be a bunch of terms: $a_1, a_2, a_3, a_4,...... $ and we can refer to each term by its index. The fifth term is $a_5$. The $39$th term is $a_{39}$ and so on.
If we pick some arbitrary term but want to indicate it might be any term we can call it be an arbitrary index. $a_n$ is the $n$-th term. It could be the $512$th term, $a_{512}$ or it could by the $7,487$th term, $a_{7,487}$. We don't know and don't care. We are refering to a non-specific term.
We are refering to a term when we don't know what position it is in. We say "let's call the position it is in: $n$".
Now, let's suppose I said "After picking that term look at the previous term" or "look at the next term". How can we refer to that. We if our term was the $a_{512}$ then the previous term would be .... $a_{511}$. And if our term was $a_{7,487}$ the next term would be .... $a_{7,488}$.
But how do we refer to it in general. If our term was $a_n$ what is the term immediately before it. Well, if this was the $n$th term, the one write before it is the $n-1$th term, $a_{n-1}$.
So in this case:
$a_n = a_{n-1} + 2$
means. "Pick any term. It is equal to the term before it plus two".
So if the previous term $a_{n-1}$ was equal to $517$ then this term, $a_n$ will be equal to $519$.
So $a_1 = 3$ because we were told so.
And $a_2 = $ the previous term plus $2 = 3+2 = 5$.
And $a_3 = a_2 + 2 = 5 + 2 = 7$ and so on...
$a_4 = a_3 + 2 = 7 + 2= 9$ .....
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
$endgroup$
$begingroup$
Thanks very much for these helpful responses. This gives me a far better understanding of where the n-1 is coming from. I appreciate the time you all took to post. Cheers
$endgroup$
– lachy
Jan 17 at 0:26
add a comment |
$begingroup$
Your sequence will be a bunch of terms: $a_1, a_2, a_3, a_4,...... $ and we can refer to each term by its index. The fifth term is $a_5$. The $39$th term is $a_{39}$ and so on.
If we pick some arbitrary term but want to indicate it might be any term we can call it be an arbitrary index. $a_n$ is the $n$-th term. It could be the $512$th term, $a_{512}$ or it could by the $7,487$th term, $a_{7,487}$. We don't know and don't care. We are refering to a non-specific term.
We are refering to a term when we don't know what position it is in. We say "let's call the position it is in: $n$".
Now, let's suppose I said "After picking that term look at the previous term" or "look at the next term". How can we refer to that. We if our term was the $a_{512}$ then the previous term would be .... $a_{511}$. And if our term was $a_{7,487}$ the next term would be .... $a_{7,488}$.
But how do we refer to it in general. If our term was $a_n$ what is the term immediately before it. Well, if this was the $n$th term, the one write before it is the $n-1$th term, $a_{n-1}$.
So in this case:
$a_n = a_{n-1} + 2$
means. "Pick any term. It is equal to the term before it plus two".
So if the previous term $a_{n-1}$ was equal to $517$ then this term, $a_n$ will be equal to $519$.
So $a_1 = 3$ because we were told so.
And $a_2 = $ the previous term plus $2 = 3+2 = 5$.
And $a_3 = a_2 + 2 = 5 + 2 = 7$ and so on...
$a_4 = a_3 + 2 = 7 + 2= 9$ .....
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
$endgroup$
Your sequence will be a bunch of terms: $a_1, a_2, a_3, a_4,...... $ and we can refer to each term by its index. The fifth term is $a_5$. The $39$th term is $a_{39}$ and so on.
If we pick some arbitrary term but want to indicate it might be any term we can call it be an arbitrary index. $a_n$ is the $n$-th term. It could be the $512$th term, $a_{512}$ or it could by the $7,487$th term, $a_{7,487}$. We don't know and don't care. We are refering to a non-specific term.
We are refering to a term when we don't know what position it is in. We say "let's call the position it is in: $n$".
Now, let's suppose I said "After picking that term look at the previous term" or "look at the next term". How can we refer to that. We if our term was the $a_{512}$ then the previous term would be .... $a_{511}$. And if our term was $a_{7,487}$ the next term would be .... $a_{7,488}$.
But how do we refer to it in general. If our term was $a_n$ what is the term immediately before it. Well, if this was the $n$th term, the one write before it is the $n-1$th term, $a_{n-1}$.
So in this case:
$a_n = a_{n-1} + 2$
means. "Pick any term. It is equal to the term before it plus two".
So if the previous term $a_{n-1}$ was equal to $517$ then this term, $a_n$ will be equal to $519$.
So $a_1 = 3$ because we were told so.
And $a_2 = $ the previous term plus $2 = 3+2 = 5$.
And $a_3 = a_2 + 2 = 5 + 2 = 7$ and so on...
$a_4 = a_3 + 2 = 7 + 2= 9$ .....
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
answered Jan 16 at 0:51
fleabloodfleablood
73.7k22891
73.7k22891
$begingroup$
Thanks very much for these helpful responses. This gives me a far better understanding of where the n-1 is coming from. I appreciate the time you all took to post. Cheers
$endgroup$
– lachy
Jan 17 at 0:26
add a comment |
$begingroup$
Thanks very much for these helpful responses. This gives me a far better understanding of where the n-1 is coming from. I appreciate the time you all took to post. Cheers
$endgroup$
– lachy
Jan 17 at 0:26
$begingroup$
Thanks very much for these helpful responses. This gives me a far better understanding of where the n-1 is coming from. I appreciate the time you all took to post. Cheers
$endgroup$
– lachy
Jan 17 at 0:26
$begingroup$
Thanks very much for these helpful responses. This gives me a far better understanding of where the n-1 is coming from. I appreciate the time you all took to post. Cheers
$endgroup$
– lachy
Jan 17 at 0:26
add a comment |
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$begingroup$
Your age and background are unimportant.
$endgroup$
– John Douma
Jan 16 at 0:31
$begingroup$
Just like we use $x$ to refer to non-specific number when we don't know which one we use $a_n$ to refer to one of the terms in some position but we don't know which position. We call it the $n$th position. So $a_{n-1}$ is the term in the position immediately before it. And $a_{n+1}$ is the term in the position immediately after it.
$endgroup$
– fleablood
Jan 16 at 0:54