To prove:
The sum of the first n even squares, less the sum of the first n squares, is equal to the nth square, plus the nth cube, plus the nth triangular number.
The sum of the first n odd squares, less the sum of the first n squares, is equal to the nth cube, less the nth triangular number.
(In reply to
re: Just the first one by Steven Lord)
D'Oh!!
I had the formula for the odd squares worked out correctly, then made a transcribing error when I moved to Notepad.
Corrected:
1st n ODD squares
(4n^3 - n)/3
Sentence Two:
(4n^3 - n)/3 - (2n^3 + 3n^2 + n)/6 = n^3 - (n^2 + n)/2
2(4n^3 - n) - (2n^3 + 3n^2 + n) = 6n^3 - 3(n^2 + n)
(8n^3 - 2n) - (2n^3 + 3n^2 + n) = 6n^3 - 3n^2 - 3n
(8n^3 - 2n^3) - (3n^2) - n - 2n = 6n^3 - 3n^2 - 3n
6n^3 - 3n^2 - 3n = 6n^3 - 3n^2 - 3n
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Posted by Larry
on 2022-11-23 11:59:25 |
re(2): Just the first one D'Oh!!
