Hard Mathematical Induction

Question

I have a mathematical induction question and I know what I need to do just not how to do it.

The question is:

Prove the equality of:

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

Base case:

$$(1 + 2)^2 = 1^3 + 2^3\\ (3)^2 = 1 + 8\\ 9 = 9$$

and I understand I have to get the sides to equal each other though I'm not sure how to do that:

I use this:

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

but i can't seem to factor anything in anyway to figure it out . . .

I've tried putting the $S(n)$ in the $S(n + 1)$:

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

but its just getting the $-(n + 1)^3$ on the first side I can't figure out...

Any help would be amazing!!!!

This has already been asked a lot of times; see eg http://math.stackexchange.com/questions/974091/prove-that-left-sum-k-1nk-right2-sum-k-1nk3-holds-true-for-n — Some Math Student, Oct 15 '14 at 11:00
See also http://math.stackexchange.com/questions/411485/prove-by-induction-that-13-23-33-n3-fracn2n124-for?rq=1, http://math.stackexchange.com/questions/974091/prove-that-left-sum-k-1nk-right2-sum-k-1nk3-holds-true-for-n, and others. — Dietrich Burde, Oct 15 '14 at 11:08

score 1 · Answer 1 · answered Oct 15 '14 at 10:56

1

The easiest way to prove this is to use the equality $$1+2+\cdots + n = \frac{n(n+1)}{2}$$

Which simplifies the left part of your equation.

answered Oct 15 '14 at 10:56

5xum

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Are you able to expand on that at all? – Michael Manning Oct 15 '14 at 10:57
@MichaelManning I am, but I think it would benefit you much more if you expanded on it instead of me. – 5xum Oct 15 '14 at 10:59
Okay so am I allowed to substitute the right hand of your statement into my lefthand side? – Michael Manning Oct 15 '14 at 11:02
I'm not in anyway asking for free answers I have spent the last three hours trying this and have written on over 15 pages – Michael Manning Oct 15 '14 at 11:02
I'll start mucking around with that for a bit – Michael Manning Oct 15 '14 at 11:03
@MichaelManning You can use the equality I wrote if you already know it to be true. Otherwise, you must prove it is true (and induction is really simple here). – 5xum Oct 15 '14 at 11:06

score 1 · Answer 2 · answered Oct 15 '14 at 11:11

Let the statement hold for $n \in \mathbb{N}$. For $n+1$, by applying the binomial formula $(a+b)^2=a^2+2ab+b^2$ you have that $$\begin{align*}\left(\underbrace{1+2+\ldots+n}_{=a}+\underbrace{(n+1)}_{=b}\right)^2&=(1+2+\ldots+n)^2+2(1+2+\ldots+n)(n+1)+(n+1)^2=\\&=1^3+2^3+\ldots+n^3+2\frac{(n+1)n}{2}(n+1)+(n+1)^2=\\&=1^3+2^3+\ldots+n^3+\underbrace{n(n+1)^2+(n+1)^2}_{=(n+1)(n+1)^2=(n+1)^3}=\\&=1^3+2^3+\ldots+n^3+(n+1)^3 \end{align*}$$

score 0 · Answer 3 · answered Oct 15 '14 at 11:05

0

Hints:

Suppose that when $n=k$, the claim holds. Now when $n=k+1$, $$(1+2+\cdots+k+(k+1)^2=(1+2+\cdots+k)^2+2(1+2+\cdots+k)(k+1)+(k+1)^2=1^3+2^3+\cdots+(k+1)^3.$$

answered Oct 15 '14 at 11:05

Paul

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did you get (1+2+⋯+k)2+2(1+2+⋯+k)(k+1)+(k+1)2 by expanding (1+2+⋯+k+(k+1)2? – Michael Manning Oct 15 '14 at 11:09
@MichaelManning: Yes. – Paul Oct 15 '14 at 11:14

Hard Mathematical Induction

3 Answers3