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

so far I have..

$$1^3+2^3+\cdots+k^3+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2$$

then..

$$\left[\frac{k(k+1)}2\right]^2+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2$$

where do I go from here so that the rhs equals the lhs?

Michael Hardy
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Lil
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1 Answers1

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Inductive step: $$\begin{align*}\underbrace{1^3+2^3+\ldots+n^3}_{=\left(\dfrac{n(n+1)}{2}\right)^2}+(n+1)^3&=\left(\dfrac{n(n+1)}{2}\right)^2+(n+1)^3=(n+1)^2\cdot\left(\frac{n^2}{4}+(n+1)\right)=\\&=(n+1)^2\cdot\left(\frac{n^2+4n+4}{4}\right)=(n+1)^2\cdot\left(\frac{(n+2)^2}{2^2}\right)=\\&=\left(\frac{(n+1)(n+2)}{2}\right)^2\end{align*}$$

Jimmy R.
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  • alright so on the left hand side I can get up to [k^2(k+1)^2+4(k+1)^3]/4 but I don't understand how the next step becomes (k+1)^2(k^2+4k+4)/4? – Lil Nov 09 '14 at 17:54
  • Factor out the term (k+1)^2 – Jimmy R. Nov 09 '14 at 17:56
  • if i factor that term I get k^2(k^2+2k+1) how will that lead to (k^2+4k+4)? – Lil Nov 09 '14 at 18:00
  • I see online people have rewritten it so that it becomes (k+1)^2((k^2/4)+k+1) but how is that equivalent to (k+1)^3? – Lil Nov 09 '14 at 18:09
  • @Lil Perhaps if you used MathJax to properly write mathematics you could see clearer what you're writing (BTW, why should anyone show that something's equivalent to $;(k+1)^3;$ ?) .Take a peek at http://meta.matheducators.stackexchange.com/questions/93/mathjax-basic-tutorial-and-quick-reference – Timbuc Nov 09 '14 at 18:14
  • Be more careful with the calculations. You did not factor out correctly. – Jimmy R. Nov 09 '14 at 18:14