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Topic review (newest first)

bob bundy
2012-12-27 02:31:31

hi therussequilibrium

Finally remembered that formula and another way to prove it.

You can use a similar method for any summation of the form

Let's say you have spotted that the formula you  want is

Start with the next highest power of n, ie.  n cubed.

Write out this expression for n=1, n=2, n=3, ...n=n

...............    ............    ..............      .............        .............

__________________________________________________________________

Now add up each column.  Most of the cube terms on the LHS cancel with most of those on the RHS.  1+1+1...+1 comes to n.

The formula for the {triangle numbers}  is

so

Re-arranging

multiply by 2 to remove the fraction and factorising out the common factor of (n+1)

and so, finally

Bob

bobbym
2012-12-22 22:46:58

Hi;

You are welcome. Have a good holiday.

therussequilibrium
2012-12-22 22:44:55

Awesome! Again thanks for your help! And I will definitely do what you suggested, I plan on devoting a lot of my time to math from here on so I'll probably be here often.

Thanks again!

bobbym
2012-12-22 22:42:10

Hi;

That is correct. Do not forget to add them up to make the formula.

therussequilibrium
2012-12-22 22:40:10

Okay, maybe it was easier than I thought to simplify, I think I figured it out. I just wasn't sure if I should distribute first or FOIL first. And what I ended up with was this:

And it seems like you can divide, which seems weird, but it comes out with the right answer.

So, it seems that I can divide like that, and I simplified correctly?

bobbym
2012-12-22 22:21:42

Hi;

Okay, I am glad to help.

About catching up, I recommend you post your questions here when you get stuck. There are knowledgeable people here.

therussequilibrium
2012-12-22 22:18:23

Yeah, I'm trying by hand, I'll get it eventually. You've given me enough help, but thank you. I do appreciate you taking your time to walk through this with me.

bobbym
2012-12-22 22:12:18

Hi;

Offhand I do not remember the derivation of that. It is from the difference calculus. But it does work.

Try simplifying it by hand. If you want to I will help but it is really not necessary.

therussequilibrium
2012-12-22 22:08:18

Well, I can't figure out how to simplify this problem. But question, where does this equation that we are using come from?

bobbym
2012-12-22 21:52:58

I am sorry, that is a factorial.

2! means 2 * 1, it equals 2

3! means 3 * 2 * 1, it equals 6

I did away with the factorials.

therussequilibrium
2012-12-22 21:47:19

hmm, that's exactly the equation I entered and that's the answer it gave me; and I'm not sure how to simplify a problem with radicals in it, on my own.

bobbym
2012-12-22 21:42:42

Hi;

That is not correct. Let me show you what I meant.

I dropped the ellipses on the end which are just there to show that there might be more terms.

When you substitute a = 0, b = 1, c = 3, d = 2 into that you get,

you can leave it like that and say

or you can simplify it,

That is what we got by the other methods. So this method solves your recurrence by just making a table and plugging into a formula.

therussequilibrium
2012-12-22 21:36:45

Not sure how to type out a fraction on the forum, but I think this is the answer:

bobbym
2012-12-22 21:20:34

Yes, the difference table now looks like this.

Take the first number in every row and say

a = 0;
b = 1
c = 3
d = 2

the rest are zero. Take those and plug into this formula.

therussequilibrium
2012-12-22 21:15:27

Correct, and the second sequence really starts at 1 and not at 4, etc.