Proof of 1^3 + 2^3 + ... n^3 = [n(n+1)/2]^2 for Positive Integers

[caws]

I am trying to prove by induction 1^3 + 2^3 + ... n^3 = [n(n+1)/2]^2
when n is a positive integer

Let P(n), if P(1) then n^3 = 1^3 = 1 and [n(n+1)/2]^2 = [1(1+1)/2]^2 = 1

the inductive hypothesis is 1^3 + 2^3 + ... k^3 = [k(k+1)/2]^2

Assuming P(k) is true then prove P(k+1) is true, insert (k+1) into problem

1^3 + 2^3 + ... k^3 + (k+1)^3 = [k+1(k+1+1)/2]^2 or [(k+1)(k+2)/2]^2

by the inductive hypothesis we get

[k(k+1)/2]^2 + (k+1)^3 = [(k+1)(k+2)/2]^2

am I thinking this through correctly? and where do I go from here?

 

[d_leet]

^2 or [(k+1)(k+2)/2]^2

by the inductive hypothesis we get

[k(k+1)/2]^2 + (k+1)^3 = [(k+1)(k+2)/2]^2

am I thinking this through correctly? and where do I go from here?

I just did this proof earlier today it's not too hard, but it looks a lot to me like you assumed what you're trying to prove what you need to do after the induction hypothesis is take the left hand side of the last equation you have and use it to arrive at the right hand side, because otherwise you have done nothing at all and just assumed the statement that you were trying to prove.

 

[caws]

Ok. I am going to try to work this through again and post tommorow.

 

[HallsofIvy]

Assuming P(k) is true then prove P(k+1) is true, insert (k+1) into problem

1^3 + 2^3 + ... k^3 + (k+1)^3 = [k+1(k+1+1)/2]^2 or [(k+1)(k+2)/2]^2

It's MUCH better to say
1^3+ 2^3+ ...+ k^3+ (k+1)^3= [k(k+1)/2]^2+ (k+1)^3 and work from there.

 

[caws]

thanks, at least I know I am on the right track and was understanding the process, now all I have to do is solve to prove.