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#1 Re: Help Me ! » Formula For the Sum Of the First N Squares » 2011-05-16 00:57:09
1^3+2^3+...+n^3+(n+1)^3 = 0^3+1^3+2^3+3^3+...+n^3+...
\______ _____/ \_______ ________/
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these terms cancel these terms .
So only the (n+1)^3 is left on the left side of the equation.
Thanks for your help!
#2 Re: Help Me ! » Formula For the Sum Of the First N Squares » 2011-05-16 00:01:19
1^3 = (0 + 1)^3 = 0^3 + 3 ( 0^2 ) + 3 (0) + 1
2^3 = (1 + 1)^3 = 1^3 + 3 ( 1^2 ) + 3 (1) + 1
3^3 = (2 + 1)^3 = 2^3 + 3 ( 2^2 ) + 3 (2) + 1
4^3 = (3 + 1)^3 = 3^3 + 3 ( 3^2 ) + 3 (3) + 1
etc.
n^3 = (n-1 + 1)^3 = (n-1)^3 + 3 (n-1)^2 + 3 (n-1) + 1(n+1)^3 = (n + 1)^3 = n^3 + 3 n^2 + 3 n + 1
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Hi bob,
Im sorry i couldnt understand
As 0 cubed is 0 the cubes on the RHS are the same as on the Left except for the one you are asking about.
im not sure which one you are referring to.
They wanted me to cancel out n^3 (in bold) but since
(n + 1)^3 = n^3 + 3 n^2 + 3 n + 1
, wouldnt cancelling n^3 makes it not balanced?
Thanks!
#3 Help Me ! » Formula For the Sum Of the First N Squares » 2011-05-15 21:42:22
- rainboi
- Replies: 5
Hi all,
I refer to "http://mathforum.org/library/drmath/view/56920.html"
I followed the steps on how I should proof 
But im stucked at
"1. All of the cubes cancel except for (n+1)^3"
I have no idea why this step is needed. Can anyone help me please? Thank you.
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