Series/Sequence Problem?

1. Apr 14, 2008

ae4jm

[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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2. Apr 15, 2008

slider142

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. Apr 15, 2008

HallsofIvy

Staff Emeritus
{an} is defined, recursively, by a1= 4, $a_{n+1}= a_n+ 4n$. 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. Apr 15, 2008

tiny-tim

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 … ?

5. Apr 15, 2008

ae4jm

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?

Last edited: Apr 15, 2008
6. Apr 15, 2008

tiny-tim

Hi ae4jm!

Humour me … what's an+2 - an? … what's an+3 - an?

7. Apr 15, 2008

ae4jm

I got 12 and 24???

8. Apr 15, 2008

tiny-tim

Hi ae4jm!

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)

9. Apr 15, 2008

ae4jm

I believe that the first one is 6-4=2 and 7-4=3, right?

10. Apr 15, 2008

ae4jm

does this look correct for
Code (Text):
$$a_{n}=4(\frac{n(n-1)}{2})+4$$

11. Apr 16, 2008

tiny-tim

Woohoo!

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

Any questions?

12. Apr 16, 2008

ae4jm

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.