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Mathmatical Induction - Hazzy - 2010-09-14
I need to prove that the sum of (2r-1)^3, r=1 to n is equal to n^2(2n^2-1) for all positive integers. I'm at a lost of what to do beyond setting it up with the Inductive Assumption and testing it with n=1.
n=1:
(2*1-1)^3 = 1^2(2*1^2-1)
1 = 1
k^2(2k^2-1) + (2(k+1)-1)^3 = (k+1)^2(2(k+1)^2-1)
Math help thread gave me the autolock thing.
Mathmatical Induction - Pysk - 2010-09-14
You are on the right track. With induction you assume that the statement is true for k and show from that that the statement is true for k+1.
The induction hypothesis gives that SUM[(2r-1)^3] from r=1 to k is equal to k^2(2k^2-1). What you need to do is show that when you add (2(k+1)-1)^3 to the right side you get (k+1)^2(2(k+1)^2-1).
Mathmatical Induction - Hazzy - 2010-09-14
I know what I need to do, not how to do it. Specifically the Algebra.
Mathmatical Induction - Pysk - 2010-09-15
Probably the easiest thing to do is to distribute out both sides. You should get that:
SUM[(2r-1)³] from r=1 to k+1 = k²(2k²-1)+(2k+1)³ = 2k^4+8k³+11k²+6k+1
and you should get that:
(k+1)²(2(k+1)²-1) = 2k^4+8k³+11k²+6k+1
and hence the two are equal.