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Prove the following statement $S(n)$ for $n\ge1$:

$$\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}{4}$$

To prove the basis, I substitute $1$ for $n$ in $S(n)$:

$$\sum_{i=1}^11^3=1=\frac{1^2(2)^2}{4}$$

Great. For the inductive step, I assume $S(n)$ to be true and prove $S(n+1)$:

$$\sum_{i=1}^{n+1}i^3=\frac{(n+1)^2(n+2)^2}{4}$$

Considering the sum on the left side:

$$\sum_{i=1}^{n+1}i^3=\sum_{i=1}^ni^3+(n+1)^3$$

I make use of $S(n)$ by substituting its right side for $\sum_{i=1}^ni^3$:

$$\sum_{i=1}^{n+1}i^3=\frac{n^2(n+1)^2}{4}+(n+1)^3$$

This is where I get a little lost. I think I expand the equation to be

$$=\frac{(n^4+2n^3+n^2)}{4}+(n+1)^3$$

but I'm not totally confident about that. Can anyone provide some guidance?

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    You are on the right path, just expand $(n+1)^2(n+2)^2$ and match the coefficients with your last formula (wherein you can also expand $(n+1)^4$).2012-10-10
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    Some inductive proofs can be found [here](http://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct). In particular [this answer](http://math.stackexchange.com/a/62198/) is along the same lines as your attempt.2012-10-10
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    You're making one of the usual mistakes: you write $\sum_{i=1}^11^3=\frac{1^2(2)^2}{4}=>1=1$. You have the arrow ($=>$) presumably meaning "if...then..." going in the wrong direction. What you need is $\sum_{i=1}^1 i^3 = 1 = \dfrac{1^2\cdot2^2}{4}$. That shows that the first expression in this string of equalities equals the third expression. This is a flaw in logical presentation that it's notoriously hard to talk students out of.2012-10-10
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    ....but that doesn't upset your later argument. At the end, you're doing OK; you just need to keep going. The common denominator is $4$, so you need $\dfrac{4(n+1)^3}{4}$ $=\dfrac{4(n^3+3n^2+3n+1)}{4}$ $\dfrac{4n^3+12n^2+12n+4}{4}$, and then add the two fractions and simplify, and then factor the numerator.2012-10-10
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    @MichaelHardy Thanks for pointing out my misused arrow. The post has been edited to fix this.2012-10-10

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