# Find the terms that add up to 1010100

1. Feb 8, 2012

### kscplay

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. Feb 8, 2012

### Joffan

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?

3. Feb 8, 2012

### kscplay

I just recopied the problem exactly as it was stated. But anyways I've already found the solution. Thanks :)

4. Feb 10, 2012

### Joffan

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?

5. Apr 6, 2012

### kscplay

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 $n^2=a_{n}+a_{n-1}$

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.

6. Apr 6, 2012

### Joffan

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

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