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Simplification of a sequence

  1. Oct 21, 2009 #1
    Hi guys,
    the x's are supposed to be *, as in 2x2 is supposed to be 2*2, it won't listen to me.
    I was wondering if it is possible to simplify this sequence:

    [tex]1)2[/tex]

    [tex]2)2^2 +4[/tex]

    [tex]3)2^3 +(2 * 4) + 6[/tex]

    [tex]4)2^4 + (2^2 * 4) + (2 * 6) +8[/tex]

    [tex]5)2^5 + (2^3 * 4) + (2^2 * 6) + (2 * 8) + 10[/tex]

    [tex]6)2^6 + (2^4 * 4) + (2^3 * 6) + (2^2 * 8) + (2 * 10) + 12[/tex]

    I've made a few attempts but there are not satisfactory:
    [tex]2^n + 2^n-2 * (n-(n-2))+2^n-3 * (n-(n-4))+...+2^2 * (2n-2) + 2^0 * 2n[/tex]
    Any help appreciated
     
  2. jcsd
  3. Oct 21, 2009 #2

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's pretty easy to do if you're able to use functions like max().

    - Warren
     
  4. Oct 22, 2009 #3
    Yeah, my attempt in Latex code didn't look how it was supposed to, I'm not particularly good with Latex. Thanks for the advice, I'll go look up Max().
     
  5. Oct 22, 2009 #4
    The only maximum function my maths textbook gives me is for trig and for calculus. Maximum values in trig don't apply here, and I can't see the application of the maximum in calculus here either. Any further advice, hints, or tips?
     
  6. Oct 22, 2009 #5

    Mark44

    Staff: Mentor

    It looks like you're on the right track. The nth term of your sequence will have n terms in it, with the first term being 2^n. The other n - 1 terms seem to follow their own pattern.

    In each of the remaining n - 1 terms, you have the product of 2 to some power and twice a number. The exponent on 2 starts at n - 2 and works its way down to 0. The "twice a number" factor starts at 2*2 and works its way up to 2*n.

    Does that help?
     
  7. Oct 22, 2009 #6
    Thanks so much for your help Mark.
    Okay, I think this is right, after some further simplification:
    [tex]2^n+2\sum_{a=2}^n a2^{n-a}[/tex]
     
    Last edited: Oct 22, 2009
  8. Oct 22, 2009 #7

    Mark44

    Staff: Mentor

    Can I assume that this is what you're looking for? Your comment about being stuck at n = 6 made me unsure.
     
  9. Oct 22, 2009 #8
    Sorry, the equation I posted with 'n' is the one I wanted. Previously I entered 6 in it's place accidently (I was reading it from my working and used one where I subtituted n with 6 to test the equation) and when I edited the Latex in the post it didn't change from 6 to n (I assumed there would be a delay of 24 hours or soemthing), but it has now. If that makes sense!
    But yes, the equation, as it appears above (with 'n'), is it correct?
     
  10. Oct 22, 2009 #9

    Mark44

    Staff: Mentor

    It looks like what I described. If your formula gives you the right values for the numbers in your sequence, I think all is good!
     
  11. Oct 22, 2009 #10
    Thanks Mark: I tried it with two (an exhaustive test), and it appears to work. It is certainly simplified! Thankyou.
     
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