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# Sigma

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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..

Mar 16, 2023

#1
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As follows: Mar 16, 2023
#2
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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
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I set j = k-2, so k=3 becomes j = 1, and k = n becomes j = n-2.  Then:

The sum $$\Sigma_{j=1}^{m}j$$  is given by $$m(m+1)/2$$,  so when m = n-2, the sum is (n-2)(n-1)/2.

Alan  Mar 16, 2023
edited by Alan  Mar 16, 2023
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Understood..thank you Alan. Stay blessed..

juriemagic  Mar 16, 2023