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

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

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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

]]>You are welcome. Have a good holiday.

]]>Thanks again!

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

]]>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?

]]>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.

]]>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.

]]>2! means 2 * 1, it equals 2

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

I did away with the factorials.

]]>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.

]]>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.

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