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.
I can prove the first sentence, but I get an error when attempting to prove the second. Perhaps I made the error.
nth triangular number
n(n+1)/2 or (n^2 + n)/2
1st n squares
[n(n+1)(2n+1)] / 6 or (2n^3 + 3n^2 + n)/6
1st n ODD squares
(4n^3 - n^2)/3 [*** <-- error is here (4n^3 - n)/3 correction ]
1st n EVEN squares
(4n^3 + 6n^2 + 2n)/3
Sentence One:
(4n^3 + 6n^2 + 2n)/3 - (2n^3 + 3n^2 + n)/6 = n^2 + n^3 + (n^2 + n)/2
2(4n^3 + 6n^2 + 2n) - (2n^3 + 3n^2 + n) = 6n^2 + 6n^3 + 3(n^2 + n)
(8n^3 + 12n^2 + 4n) - (2n^3 + 3n^2 + n) = 6n^2 + 6n^3 + 3n^2 + 3n
(8n^3 - 2n^3) + (12n^2 - 3n^2) + (4n - n) = 6n^3 + (6n^2 + 3n^2) + 3n
Sentence Two:
(4n^3 - n^2)/3 - (2n^3 + 3n^2 + n)/6 = n^3 - (n^2 + n)/2
2(4n^3 - n^2) - (2n^3 + 3n^2 + n) = 6n^3 - 3(n^2 + n)
(8n^3 - 2n^2) - (2n^3 + 3n^2 + n) = 6n^3 - 3n^2 - 3n
(8n^3 - 2n^3) - (2n^2 + 3n^2) - n = 6n^3 - 3n^2 - 3n
6n^3 - 5n^2 - n is not equal to 6n^3 - 3n^2 - 3n
Edited on November 23, 2022, 12:00 pm
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Posted by Larry
on 2022-11-23 10:51:41 |
Just the first one
