SOLUTION: Proof that the sums of the first N Square Numbers is given by =[n(n+1)(2n+1)]/6

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Question 1057029: Proof that the sums of the first N Square Numbers is given by
=[n(n+1)(2n+1)]/6
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Proof that the sums of the first N Square Numbers is given by
=[n(n+1)(2n+1)]/6
---
Show it's true for n = 1::
1 = [1(2)(3)]/6
1 = 1
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Assume it's true for n = k::
1 + 4 + 9 +...k^2 = [k(k+1)(2k+1)]/6
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Show it is true for n = k+1
---
1 + 4 + 9 + .. k^2 + (k+1)^2
----
= [k(k+1)(2k+1)]/6 + (k+1)^2
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= [(k+1)[k(2k+1)]/6+(k+1)]
-------
= [(k+1)[2k^2 + k + 6k + 6]/6
= [(k+1)(2k^2 + 7k + 6]/6
= [(k+1)[(2k^2+ 4k + 3k + 6]/6
-----
= [(k+1)[2k(k+2)+3(k+2)]/6
= [(k+1)[(2k+3)(k+2)]/6
= [(k+1)(2(k+1)+1)((k+1)+1]/6
----
= [(k+1)((k+1)+1)(2(k+1)+1]/6
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Cheers,
Stan H.
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