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I'm sitting with the proof in front of me, but I do not understand it.

$$A = \{n \in Z^{++} \mid (1^3 + 2^3 + 3^3+\cdots+n^3) = (1 + 2 + \cdots + n)^2\}$$

The first step of proof by induction is simple enough,to prove that $1 \in A$

$1^3 = 1^2$

The next step is where I get tripped up. So I add $n + 1$ to the right hand side

$$(1 + 2 + \cdots + n + (n + 1))^2 = (1 + 2 + \cdots + n)^2 + 2(1 + 2 + \cdots + n)(n + 1) + (n + 1)^2$$

My algebra is failing me here, because I do not understand how the equation was expanded.

Michael Hardy
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Leon
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    Unless I'm mistaken, that should be $n + 1$ and not $k + 1$. And the algebra you're missing is: $(a + b)^2 = a^2 + 2ab + b^2$. – M. Vinay Jun 13 '14 at 11:14
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    The expansion is $(a+b)^2=a^2+2ab+b^2$ where $a=1+2+\dots +n$ and $b=n+1$ [you have $k$ instead of $n$ on the right hand side of your final equation in the terms which I have denoted by $b$] – Mark Bennet Jun 13 '14 at 11:14
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    "Adding $n+1$ to RHS" is bad terminology, that's not what you're doing. What you're doing is taking $n\in A$ and you're wishing to prove that $n+1\in A$, so you'll be trying to prove that $$(1+2+\ldots+n+(n+1))^2=1+2^3+\ldots + n^3+(n+1)^3.$$ To do this you can take one of the sides of the equality and somehow, by a chain of equalities, reach the other one. This is what's being done. But there is typo, it should be $n$ wherever you have $k$. – Git Gud Jun 13 '14 at 11:16

2 Answers2

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For the inductive step:

$$(\underbrace{1+2\cdots+n}_{=A=\frac{n(n+1)}{2}}+(n+1))^2=\underbrace{A^2}_{=1^3+2^3+\cdots+n^3}+\underbrace{(n+1)^2+n(n+1)^2}_{=(n+1)^3}$$

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The last step uses the identity: $$(a + b)^2 = a ^2 + 2ab + b^2$$

So: $$(\underbrace{1 + 2 + \cdots + n}_a + \underbrace{(n + 1)}_b)^2 =\\= (1 + 2 + \cdots + n)^2 + 2(1 + 2 + \cdots + n)(n+1) + (n + 1)^2\tag{1}$$

By induction hypotesis we have that $(1 + 2 + \cdots + n) ^2 = (1^3 + 2^3 + \cdots + n^3)$. We also use another identity, namely $(1 + 2 + \cdots + n) = n(n+1)/2$.

Combining those two facts and substituting into $(1)$: $$(1^3 + 2^3 + \cdots + n^3) + 2\times n(n+1)/2 \times (n + 1) + (n+1)^2 = \\(1^3 + 2^3 + \cdots + n^3) + n(n+1)^2 + (n+1)^2 =\\1^3 + 2^3 + \cdots + n^3 + (n + 1)^3$$

$\blacksquare$

rubik
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