Find the terms that add up to 1010100

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The discussion revolves around finding terms from the sequence defined by triangular numbers that sum to 1010100. Participants note that the sequence consists of sums of integers up to n, with the formula a_n = n(n+1)/2. One user successfully identifies a solution involving two triangular numbers, while another shares their method of breaking down 1010100 into smaller components and proving relationships between the numbers. The use of Excel for calculations is also mentioned as a practical approach. Overall, the conversation highlights different strategies for solving the problem involving triangular numbers.
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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.
 
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Can you post a_1, a_2, a_3, a_4, ..., a_{10}? Just to show you're taking this problem seriously...

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

... a_{899}+a_{1100}=1010100 ...

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

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

btw sorry for the late response.
 
That's a neat use of pattern, well done.

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

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