Deriving Summation Formulas for the Sum of Consecutive Powers - Part 3 (k=2)

Course Web Page: sites.google.com/view/slcmath...
Summation Formula - Approach 1: drive.google.com/file/d/12grS...
Summation Formula - Approach 2: drive.google.com/file/d/12iI2...
Deep Dive - Approach 1: drive.google.com/file/d/12nK6...
Deep Dive - Approach 2: drive.google.com/file/d/12mnr...

[@Okosh630]

amazing video!!!

[@gngeannakakes]

This geometric route is very intuitive. Is there an algebraic route too? When I try Sum[1,n] i^2 = (n-0)^2 + (n-1)^2 + (n-2)^2 + ... + 3^2 + 2^2 + 1^2 = Sum[1,n] (n^2 - 2(i-1)n + (i-1)^2) I end up with nonsense even though Sum[1,n] i = (n-0) + (n-1) + (n-2) + ... + 3 + 2 + 1 = Sum (n - (i-1)) clearly leads to an answer.

[@slcmathpc]

The trick that you are mentioning indeed only works when the power is one, so it does not generalize to larger powers. I have added the links to four documents on the subject in the description of this video; your question is fully answered in the document "Summation Formula - Approach 2". Since you appear to be curious, I encourage you to check out the "Deep Dive" documents where I derive the ultimate result that allows one to handle Riemann sums in the most efficient way possible by completely bypassing the need for summation formulas! Have fun. :-)

[@Knowledgeissuccess486]

But i still can't find the sum of (n-1) where i= 1 to n. Pls help🥺

[@slcmathpc]

Well, if I get your question correctly, then you are simply summing (n-1), which is constant with respect to i, n-times, so the answer to your sum is n(n-1). :-)