Recursive sequense

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Darya

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A sequence is defined recursively: a(1)=2, a(n+1)=a(n)+2n
The task is to find a(100)
Is there a way I translate this form of sequence to the form with n only? I mean I've tried but with this particular sequence, it's not that obvious what the pattern is. Thanks!!
 
Darya said:
A sequence is defined recursively: a(1)=2, a(n+1)=a(n)+2n
The task is to find a(100)
Is there a way I translate this form of sequence to the form with n only? I mean I've tried but with this particular sequence, it's not that obvious what the pattern is. Thanks!!
Click to expand...

Yes, there is a method for getting the closed form for a linear inhomogeneous difference equation of the form you've posted. Once you've responded to the hint given above (I don't want to trample help already given), I will demonstrate this method. It may not be familiar to you unless you've studied ordinary differential equations.
 
Are these correct?
\(\displaystyle \bf{a_1=2,~a_2=4 ,~a_3=8 ,~a_4=14}~?\)
 
Darya said:
A sequence is defined recursively: a(1)=2, a(n+1)=a(n)+2n
The task is to find a(100)
Is there a way I translate this form of sequence to the form with n only? I mean I've tried but with this particular sequence, it's not that obvious what the pattern is. Thanks!!
Click to expand...
lev888 said:
Post the first few members of the sequence (don't simplify).
Click to expand...
I'll demonstrate what lev888 presumably means; it's a very good hint.

a(1) = 2
a(2) = a(1) + 2*1 = 2 + 2*1

Do you follow that? I did no simplification, just substituted the expression for one term into the formula for the next.

Do the same for two more terms. You'll find that if you fully evaluate each term, you'll agree with pka; but if you stare at the unsimplified form for a(4), you'll recognize most of the calculation from your study of series (assuming you have learned the basics).
 
Dr.Peterson said:
I'll demonstrate what lev888 presumably means; it's a very good hint.

a(1) = 2
a(2) = a(1) + 2*1 = 2 + 2*1

Do you follow that? I did no simplification, just substituted the expression for one term into the formula for the next.

Do the same for two more terms. You'll find that if you fully evaluate each term, you'll agree with pka; but if you stare at the unsimplified form for a(4), you'll recognize most of the calculation from your study of series (assuming you have learned the basics).
Click to expand...
Thank you for your answer and excuse me for the late reply. I did 6 terms without simplification. It seems like the terms 2-4 follow the pattern 2^(n) + 2n but then it doesn't fit. Is there some easy way to derive it?
 
Darya said:
I did 6 terms without simplification. It seems like the terms 2-4 follow the pattern 2^(n) + 2n but then it doesn't fit. Is there some easy way to derive it?
Click to expand...

Please show how you came to this conclusion. The formula will not be exponential. What I am hoping you will recognize is that the sequence consists of partial sums of an arithmetic sequence. But I can't help you see that until you do what we asked, and SHOW us the terms you wrote. You are probably doing something wrong, but we can't tell what without seeing your work.
 
Dr.Peterson said:
Please show how you came to this conclusion. The formula will not be exponential. What I am hoping you will recognize is that the sequence consists of partial sums of an arithmetic sequence. But I can't help you see that until you do what we asked, and SHOW us the terms you wrote. You are probably doing something wrong, but we can't tell what without seeing your work.
Click to expand...
a(2)=2+2*1
a(3)=4+2*2
a(4)=8+2*3
a(5)=14+2*4

As I said, a(2-4) seem to follow the pattern. Problems with recursive sequences are new to me and in my classes, we didn't have such at all. I figure that math in schools of different countries someway really differs.
 
Darya said:
a(2)=2+2*1
a(3)=4+2*2
a(4)=8+2*3
a(5)=14+2*4

As I said, a(2-4) seem to follow the pattern. Problems with recursive sequences are new to me and in my classes, we didn't have such at all. I figure that math in schools of different countries someway really differs.
Click to expand...
Please write out the expressions without doing any calculations.
 
lev888 said:
Please write out the expressions without doing any calculations.
Click to expand...
a(1)=2 = 2+2*0
a(2)=2+2*1 =2+1*2
a(3)=2+2*1+2*2 =2+3*2
a(4)=2+2*1+2*2+2*3 = 2+6*2
a(5)=2+2*1+2*2+2*3+2*4 =2+10*2

I expect it to be 2+smth, no??
 
Just look at the unsimplified forms:

a(1)=2
a(2)=2+2*1
a(3)=2+2*1+2*2
a(4)=2+2*1+2*2+2*3
a(5)=2+2*1+2*2+2*3+2*4

and so on. I mentioned partial sums of an arithmetic series as a very explicit hint (though I accidentally typed "sequence"). Have you learned anything about that?
 
Dr.Peterson said:
Just look at the unsimplified forms:

a(1)=2
a(2)=2+2*1
a(3)=2+2*1+2*2
a(4)=2+2*1+2*2+2*3
a(5)=2+2*1+2*2+2*3+2*4

and so on. I mentioned partial sums of an arithmetic series as a very explicit hint (though I accidentally typed "sequence"). Have you learned anything about that?
Click to expand...
Omgggg *facepalm*
Honestly, it's a whole miracle it took me so long to realize it.
Then I don't even need the n form of it. The task is to find a(100) so it's 9902.
 
Dr.Peterson said:
Just look at the unsimplified forms:

a(1)=2
a(2)=2+2*1
a(3)=2+2*1+2*2
a(4)=2+2*1+2*2+2*3
a(5)=2+2*1+2*2+2*3+2*4

and so on. I mentioned partial sums of an arithmetic series as a very explicit hint (though I accidentally typed "sequence"). Have you learned anything about that?
Click to expand...
thank you so much!!! I hope you're having a nice day!
 
Darya said:
A sequence is defined recursively: a(1)=2, a(n+1)=a(n)+2n
The task is to find a(100)
Is there a way I translate this form of sequence to the form with n only? I mean I've tried but with this particular sequence, it's not that obvious what the pattern is. Thanks!!
Click to expand...

I would express the recursion as the linear inhomogeneous difference equation:

[MATH]a_{n+1}-a_n=2n[/MATH]
The homogeneous solution will take the form:

[MATH]h_n=c_1[/MATH]
And so the particular solution must take the form:

[MATH]p_n=n(An+B)[/MATH]
Substituting into the difference equation, we get:

[MATH](n+1)(A(n+1)+B)-n(An+B)=2n[/MATH]
[MATH](2A)n+(A+B)=(2)n+0[/MATH]
Equating like coefficients, we obtain the system:

[MATH]2A=2\implies A=1[/MATH]
[MATH]A+B=0\implies B=-1[/MATH]
And so our particular solution is:

[MATH]p_n=n(n-1)[/MATH]
Hence the general solution, by the principle of superposition, is:

[MATH]a_n=h_n+p_n=c_1+n(n-1)[/MATH]
We may now use the given initial value to determine the value of the parameter:

[MATH]a_1=c_1=2[/MATH]
And so the closed form for the given recursion is:

[MATH]a_n=n(n-1)+2[/MATH]
 
By the way...sequence, not sequense. ;)

It's always good to "see a light go on" in someone's head when you know they've been struggling with the concept.
Now that you've solved this one, you'll hopefully have much less trouble with others like it when they are posed to you.
 
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