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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 $(n+1)^4$). –  filmor Oct 10 '12 at 15:04
    
Some inductive proofs can be found here. In particular this answer is along the same lines as your attempt. –  Martin Sleziak Oct 10 '12 at 15:05
    
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. –  Michael Hardy Oct 10 '12 at 16:00
    
....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. –  Michael Hardy Oct 10 '12 at 16:06
    
@MichaelHardy Thanks for pointing out my misused arrow. The post has been edited to fix this. –  Brian Oct 10 '12 at 19:30
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2 Answers 2

Note that

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

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+1 I don't know why so many people want to expand the $(n+1)^3$ when this approach is far clearer. –  Peter Taylor Oct 10 '12 at 19:42
    
Especially when you want the final form to have an $(n+1)^2$ factor. Factoring it out from the start just makes sense. –  Mike Oct 10 '12 at 20:04
    
My algebra is rusty and the intellectual leap to $\frac{(n+1)^2}{4}[n^2+4(n+1)]$ in the first equality escapes me. Can you explain further? –  Brian Oct 10 '12 at 20:39
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So far you are doing fine. Now expand $(n+1)^3$ to a polynomial, combine it with the $\frac14 (n^4+2n^3+n^2)$ term, and check to see if the result is what you expect. What you expect should be $n^2(n+1)^2\over 4$, except with $n+1$ in place of $n$.

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