Okay,

Here is a new problem that needs worked on.  Let's see what everyone comes up with.  This is a discussion that has branched off of an earlier thread.  If you want to read up on it, go here.

Here is a copy of my last post.  Let's see what transpires!

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Here is what I came up with so far:

Something you should know is that the sum of consecutive triangular numbers is actually the larger number squared.  So for example, the sum of the 5th and 6th triangular numbers is 36 or 6^2.  So let's take this idea for working sum of the nth triangular numbers.

1st + 2nd + 3rd + ... + nth = (1st + 2nd) + (3rd + 4th) + ... + ((n-1)th + nth)

*NOTE: First off let's assume n is even.

This will become 2^2 + 4^2 + 6^2 + ... + n^2.

So how do we break this up?  Let's factor out a 2^2 or a 4 and get:

4 x [1^2 + 2^2 + 3^2 + ... + (n/2)^2] = 4 x (a)(a+1)(2a+1)/6 with a = n/2  *See sequences

So now we have a basic formula that will work as long as we are looking for an even sum. 

Now, what I need your help on is if it is odd.  The only thing that I can come up with is finding the sum of the sequences as down above and add in the nth triangular number.  So for our example we are going to have the sum of the first 7 triangular numbers.  In doing this, we take the sum of the first 6 triangular numbers and get:

4 x 3(4)(7)/6 = 4 x 14 = 56.  Now we need to add the 7th triangular number which is 7(8)/2 or 28.  So 28 + 56 = 84.  I know though this is not the easiest way of doing this.  There HAS to be a simpler formula that will work.  See what you guys can come up with.

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