Series/Sequence Problem?

  • Thread starter ae4jm
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  • #1
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[SOLVED] Series/Sequence Problem?

I'm trying to figure out a formula for this sequence problem, rather than doing this over and over 1,000 times. Does anyone have a clue for the formula to find this? I've pasted the info and also the answer.

Thanks for your time!
 

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Answers and Replies

  • #2
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Write out the formula for a1, a2, and a3 without summing and find the pattern so that you can write a formula for an that does not refer to an-1. It should remind you of a simple type of sum you already know a shortcut for.
 
  • #3
HallsofIvy
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{an} is defined, recursively, by a1= 4, [itex]a_{n+1}= a_n+ 4n[/itex]. Find a1000.

The first thing I would do is start calculating a few values (hoping I won't have to go up to 1000!).

a2= 4+ 4(1), a3[/sup]= 4+ 4(1)+ 4(2), a4[/sup]= 4+ 4(1)+ 4(2)+ 4(3).

Hmmm, looks to me like an= 4(1+ 2+ 3+ ...(n-1)) so a1000= 4(1+ 2+ 3+ ... + 999). Can you find 1+ 2+ 3+ ...+ 999? It's an arithmetic sequence with common difference 1. Or you could use "Euler's method".
 
  • #4
tiny-tim
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Another method: rewrite the equation as:

an+1 - an = 4n;
so an+2 - an+1 = 4(n+1).​

Add them … what do you get … ? And then … ? :smile:
 
  • #5
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Gentlemen, I'm totally stuck. I sat here for the last 1.5 hrs and tried to figure this one out. I think that I'm making it too difficult. I've looked at the sequence all the way up through a6 and I'm just not seeing the formula.

I do know that to find the sum from 1 to 1000, would be (1000(1001))/2 or (n(n+1))/2. But I've tried finding the an like this also...dang I'm making this tuff aren't I?
 
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  • #6
tiny-tim
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Hi ae4jm! :smile:

Humour me … what's an+2 - an? … what's an+3 - an? :smile:
 
  • #7
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I got 12 and 24???
 
  • #8
tiny-tim
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Hi ae4jm! :smile:

Yes … that's a3 - a1 and a4 - a1.

But you'll get more of an idea of what's going on if you answer the general question:

what's an+2 - an? … what's an+3 - an? (in terms of n):smile:
 
  • #9
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I believe that the first one is 6-4=2 and 7-4=3, right?
 
  • #10
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does this look correct for
Code:
[tex]a_{n}=4(\frac{n(n-1)}{2})+4[/tex]
 
  • #11
tiny-tim
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:biggrin: Woohoo! :biggrin:

And now, of course, simplify it to 2n(n - 1) + 4.

Any questions? :smile:
 
  • #12
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Thanks! That was a dandy, for me anyways! I'm sure this practice will help me recognize these sequences a little better in the future. Thanks for sticking in there with me.
 

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