Find the terms that add up to 1010100

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Homework Help Overview

The problem involves identifying which terms from a sequence of triangular numbers sum to 1010100. The sequence is defined as {an}∞n=1 = {n(n+1)/2}∞n=1, representing the sums of the first n natural numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the properties of triangular numbers and their relationship to the target sum. Some suggest finding specific terms of the sequence, while others explore patterns and approximations related to the sequence.

Discussion Status

There is a mix of attempts to clarify the problem and share insights. Some participants have indicated they have found solutions, while others are still exploring different methods and approaches. The discussion includes various interpretations of how to combine terms to reach the target sum.

Contextual Notes

One participant mentions a specific method involving separating the target sum into components related to triangular numbers, while another notes the use of computational tools for exploration. There is a lack of consensus on the best approach, with multiple strategies being discussed.

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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 [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?
 
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, [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?
 
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 [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.
 
That's a neat use of pattern, well done.

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

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