Proof of the sum of r^2 series

How the proof the formula for the sum of the first n r^2 terms.

[@xaxuser5033]

Nice proof

[@WhiteboardMaths]

Thanks:)

[@EE-Spectrum]

Your explanation is superb. I understand all that you did even though I am not so good at math. The only problem is why did you start out by saying consider the summation from 1 to n of (r+1)^3-r^3?

[@nosir1479]

It's because that summation allows for this method of cancelling all the middle terms, leaving you with two terms to work with. It's called the method of differences and you'll find that with certain summations, you are able to find patterns, and cancel terms, leaving you with something simpler to work with,

[@Chaudharys1]

Where did the second line come from?

[@itzmaisie2806]

Nice one

[@WhiteboardMaths]

Thanks

[@johnmomoh4574]

Now I understand

[@MorbidlyObeseCat]

Very good

[@breadzeppelin2705]

Thank you so much sir.

[@jamessharrock278]

Where do you get the second line?

[@angelmendez-rivera351]

The second line is just a standard telescoping sum to think about. The reason you choose to telescope over r^3 is because r^3 is one power above r^2, so the cubic monomials will cancel to 0, and leave only quadratic and linear polynomials, which is what you want. If you wanted the sum of r^n, then you would telescope with (r + 1)^n - r^n. This is just the obvious choice.

[@senpaichristian7923]

What's the summation formula for (x²+1)/(x²-1) please help solve

[@joshuaford2492]

I'm sure you probably don't care anymore, but if you do: First of all it will have to be the sum from x=2 otherwise you'll divide by 0 which maths doesn't like. Next you can rewrite the sum as the sum from x=2 to n of 2/(x^2 -1) - 1 (I stumbled across it but they are equivalent) You can then use difference of two squares, then partial fractions to seperate the sum further this can be evaluated using method of differences and also the "1" becomes n-1 when summed from x=2 to n this all comes to: 1/2 -1/n -1/(n+1) + n Wow I must be really bored

[@joshuaford2492]

sorry i meant to type 2/(x^2-1) + 1

[@pagmanproductions1686]

you have pi thousand subs

[@everythingwekhow]

2r-1

[@KartikP-ob5lh]

Noice

[@bassam88888]

Nice proof