
#1
Feb2508, 06:59 PM

P: 22

1. The problem statement, all variables and given/known data
Use induction to prove the equation (1+2+...+n)^2 = 1^3 + 2^3 + ... + n^3 2. The attempt at a solution I've done three inductive proofs previous to this one where I showed that the equation was true for some case (usually n=1), then assumed it was true for n, and proved it was true for n+1. This proof doesn't seem to fit that form though because for the other proofs I could write it as the right hand side of the equation (since it's assumed that it is true for n) plus the right hand side of the equation when n+1 is plugged in. But in this case.. (1+2+...+n+(n+1))^2 != (1+2+...+n)^2 + (n+1)^2 So, I guess I haven't done a proof like this and could use a nudge in the right direction.. 



#2
Feb2508, 07:02 PM

P: 206

What is [tex]\sum^{n}_{i=1} i[/tex]?




#3
Feb2508, 07:11 PM

P: 22

n(n+1)/2
In an effort to see where you're going with this, I thought I'd do the same for i^2 and got n^2(n^2 + 1)/4 Either this isn't correct or I'm not sure where you're headed. 



#4
Feb2508, 07:16 PM

P: 206

Inductive Proof on Well Known Sum 



#5
Feb2508, 07:27 PM

P: 22

Alright, so the left hand side is equal to ((n(n+1))/2)^2 or (n^4 +2n^3 +n^2)/4...




#6
Feb2508, 07:31 PM

P: 206

So prove that (n(n+1)/2)^2 is equal to the right hand side. Much easier to verify.




#7
Feb2508, 09:33 PM

P: 22

That is a proof of the equality but it isn't inductive though. Inductive proofs always follow the form:
1) prove the base case 2) write the induction hypothesis for the case of n (assume its true basically) 3) use 2) to prove for n+1 



#8
Feb2508, 09:54 PM

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#9
Feb2508, 10:02 PM

P: 22

You're talking about the binomial theorem right?



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