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(1+2+3+.......+(n-1))^2 + n^3 = (1+2+3+.......+n)^2 Is this correct?

Posted on 2011-02-11
Medium Priority
Last Modified: 2012-06-22
I have seen somewhere that this is true,  can somebody help me in proving this?
Question by:mohet01
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LVL 31

Accepted Solution

GwynforWeb earned 260 total points
ID: 34877235

(1+2+3+.......+(n-1))^2 + n^3

= (n(n-1)/2)^2 + n^3  

= (n^4 -2n^3 +n^2)/4 + n^3  

= (n^4 +2n^3 +n^2)/4


= (n(n+1)/2)^2

= (1+2+3+.......+n)^2

yes it is true

Assisted Solution

shadow77 earned 240 total points
ID: 34877256
Let k = 1+2+3 +...+(n-1).  Then, k = (n-1)*(1 + n-1)/2 [# of terms times average value].
And k = (n^2 - n)/2.

Left hand side (lhs) = k^2 +n^3.
lhs = ((n^2 - n)/2)^2 + n^3.
lhs = (n^4 - 2n^3 + n^2)/4 + n^3.
lhs = (n^4 - 2n^3 + n^2 + 4n^3)/4.
lhs = (n^4 + 2n^3 + n^2)/4.

Right hand side (rhs) = (k + n)^2
rhs = ((n^2 - n)/2 + n)^2.
rhs = ((n^2 - n + 2n)/2)^2.
rhs = ((n^2 + n)/2)^2.
rhs = (n^4 + 2n^3 + n^2)/4.

Thus, lhs = rhs.

Author Closing Comment

ID: 34878093

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