Do a little exploration.
1^2 = 1^3 -- That's good.
(1+2)^2 = 2^3 -- That's good.
What's really going on at each step? Let's just try some arbitrary values. What the difference between (a+b)^2 and (a+b+c)^2, for example?
(a+b)^2 = a^2 + 2ab + b^2
(a+b + c)^2 = a^2 + 2ab + b^2 + 2ac + 2bc +c^2 -- Well. that looks like we can do something with it. We added the new term, multiplied by twice each previous term and added a new sqaured term. Let's see if it works again.
(a+b + c + d)^2 = a^2 + 2ab + b^2 + 2ac + 2bc +c^2 + 2ad + 2bd + 2cd + d^2 -- I like it. It's like a cute little lemma. Can we borrow from Spanish and call it a lemmita?
I think this will solve the problem for us, but first, let's restate it a little.
Given the wonderfully wistful assumption
(1+2+3+...+n)^2 = 1^3 + 2^3 + 3^3 + ... + n^3
We wish to show that
(1+2+3+...+n +(n+1))^2 = 1^3 + 2^3 + 3^3 + ... + n^3 + (n+1)^3
Or, using our wistful assumption and our new little "new term, multiplied by twice each previous term and added a new sqaured term" lemmita, what is needed for proof is now:
2*1*(n+1) + 2*2*(n+1) + 2*2*(n+1) + ... + 2*n*(n+1) + (n+1)^2 = (n+1)^3
You need to believe this.
It's pretty obvious that (n+1) is a factor of everything, so we have the simpler expression to prove:
2*1 + 2*2 + 2*2 + ... + 2*n + (n+1) = (n+1)^2
You need to believe this.
The last thing we have going for us is that almost common 2. Let's factor that out a little.
2*(1 + 2 + 3 + ... + n) + (n+1) = (n+1)^2
I see it, now. We know the sum of the first n integers.
\(\displaystyle 2\cdot\frac{n\cdot(n+1)}{2}\;+\;(n+1)\;=\;(n+1)^2\)
Simplify a little.
\(\displaystyle n\cdot(n+1)\;+\;(n+1)\;=\;(n+1)^2\)
Pretty obviously, again, (n+1) is a common factor.
Can you finish? Do you see it?
Note: I didn't like the ellipses on both sides, either. What did I do? I threw one out with the lemmita! This is called mathematics. If you don't have the tools you need, create them! If you don't like the question that is being asked, find a way to create an equivalent question that you like better!
Note: I have no doubt there are other ways to solve this one. It's just what came into my head first. It's probably important to know that I don't recall seeing this particular problem, before. That upfront exploration was very important. Don't be afraid to get your hands dirty.
