- #1

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## Homework Statement

Prove the formula by induction:

[tex]1^3 + ... + n^3 = (1 + ... + n)^2[/tex]

## Homework Equations

[tex]1 + ... + n = n(n+1)/2[/tex]

## The Attempt at a Solution

I started by showing that the formula holds for 1, since:

[tex]1^3 = (1)^2 = 1[/tex]

Then I set about trying to prove that, if [tex]n[/tex] is true then [tex]n+1[/tex] is true as well, since this would prove the formula since we have our first validation of [tex]n[/tex] in the correct formula for [tex]1^3[/tex]. Here goes:

[tex] 1^3 + ... + n^3 = (1 + ... + n)^2[/tex]

[tex] 1^3 + ... + n^3 + (n+1)^3 = (1 + ... + n)^2 + (n + 1)^3[/tex]

Now I'll work with [tex](1 + ... + n)^2 + (n + 1)^3[/tex], since that is equivalent to [tex] 1^3 + ... + n^3 + (n + 1)^3[/tex].

I'm trying to prove, then, that [tex](1 + ... + n)^2 + (n + 1)^3 = (1 + ... + n + n + 1)^2[/tex].

[tex](1 + ... + n)^2 + (n + 1)^3 = (1 + ... + n + n + 1)^2[/tex]

[tex](n + 1)^3 = (1 + ... + n + n + 1)^2 - (1 + ... + n)^2[/tex]

[tex](n + 1)^3 = 2(n + 1)(1 + ... + n) + (n + 1)^2[/tex]

Next I used the relevant equation:

[tex](n + 1)^3 = 2(n + 1)(n(n + 1)/2) + (n + 1)^2[/tex]

Now it's just basic algebra:

[tex](n + 1)^3 = (n + 1)(n)(n + 1) + (n + 1)^2[/tex]

[tex](n + 1)(n+1)(n+1) = (n+1)(n^2 + n) + (n^2 + 2n + 1)[/tex]

[tex](n^2 + 2n + 1)(n + 1) = (n^3 + 2n^2 + n) + (n^2 + 2n + 1)[/tex]

[tex] n^3 + 3n^2 + 3n + 1 = n^3 + 3n^2 + 3n + 1[/tex]

Therefore:

[tex](1 + ... + n)^2 + (n + 1)^3 = (1 + ... + n + n + 1)^2[/tex].

Therefore:

[tex](1 + ... + n + n + 1)^2 = (1^3 + ... + n^3 + (n+1)^3)[/tex].

So what is true for [tex]n[/tex] is true for [tex]n + 1[/tex] and the proof by induction seems to be complete. I just learned this method of proof, so critiques appreciated. Sorry for how long it is! I have a feeling it could be cleaned up more than a little bit.

Thanks.