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Find the terms that add up to 1010100

  • Thread starter kscplay
  • Start date
  • #1
23
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Homework Statement


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

Homework Equations



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.
 

Answers and Replies

  • #2
473
13
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?
 
  • #3
23
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I just recopied the problem exactly as it was stated. But anyways I've already found the solution. Thanks :)
 
  • #4
473
13
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?
 
  • #5
23
0
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.
 
  • #6
473
13
That's a neat use of pattern, well done.

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

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