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Find the terms that add up to 1010100

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Which terms of this sequence add up to 1010100? (Don't need to be consecutive terms)

    {an}n=1 = {n(n+1)/2}n=1
    2. Relevant equations

    3. The attempt at a solution
    The sequence is made up of the sums of all the numbers less than and including n. Don't really know much more than that.
     
  2. jcsd
  3. Feb 8, 2012 #2
    Can you post [itex]a_1, a_2, a_3, a_4, ..., a_{10}[/itex]? Just to show you're taking this problem seriously...

    Now, for large n, [itex]a_n\approx n^2/2[/itex]. Can you find the largest [itex]a_n[/itex] less than the target value?
     
  4. Feb 8, 2012 #3
    I just recopied the problem exactly as it was stated. But anyways I've already found the solution. Thanks :)
     
  5. Feb 10, 2012 #4
    No problem. There is actually one solution that involves adding just two triangular numbers together, [itex]a_p+a_q=1010100[/itex].

    ... [itex]a_{899}+a_{1100}=1010100[/itex] ...

    Just as a matter of interest, how did you solve it?
     
  6. Apr 6, 2012 #5
    You're solution is much shorter than mine. How did you do it? I don't know much about triangular numbers. I separated 1010100 in to = 10002 + 1002 + 102. I noticed the pattern that [itex]n^2=a_{n}+a_{n-1}[/itex]

    So then I proved that: [itex][\frac{n(n+1)}{2}]+[\frac{(n-1)n}{2}]=n^2[/itex]
    Afterwards, it was easy: [tex]1000^2+100^2+10^2= (a_{1000}+a_{999}+a_{100}+a_{99}+a_{10}+a_{9})[/tex]

    btw sorry for the late response.
     
  7. Apr 6, 2012 #6
    That's a neat use of pattern, well done.

    I used my standard "engineering" approach: Excel :-).
     
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