Hi friends,

Trust you are all doing well...please, i really have only this last question....I have tried several things...but just do not get it...

Prove that \(\sum_{k=3}^n (2k-1)n=n^3-4n\)

If I evaluate the left hand side, I end up with \(Tn=2n^2-3n\)

So I was thinking that to prove \(2n^2-3n = n^3-4n\), I could just subtract the one from the other, and I should get 0..alas, this is not working...I am asking please for some guidance here...Thank you for your time..

juriemagic Mar 16, 2023

#1

#2**0 **

Hi Alan,

Gosh, what a piece of work!!..Thank you very much indeed.

I follow the sum up to the point where \(2j +3\) changes to \(j+3(n-2)\),

Could I ask if you would please just explain that bit to me?...Thanks Alan...

juriemagic
Mar 16, 2023