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Can anyone show me how to simplify this fraction:

$$ \frac{(k + 1)((k + 1)+1)(2(k + 1)+1)}{6}\;\;. $$

What can be factored out and so forth?

Thanks.

user1176409
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Cfrim
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  • http://math.stackexchange.com/questions/183316/how-to-get-to-the-formula-for-the-sum-of-squares-of-first-n-numbers – lab bhattacharjee Apr 29 '14 at 16:40
  • This seems to be equal to some other fraction involving a power of $k$, but I don't remember the exact form. The best way to show that it can be done is probably either using an induction proof or possibly multiplying everything out – Maroon Apr 29 '14 at 16:41

2 Answers2

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You can simplify it to $\frac{(k+1)(k+2)(2k+3)}6$ and that's it. While we do know that one of the consecutive integers $k+1$ and $k+2$ must be even and hence somehow a factor of $2$ can be cancelled, we cannot know which of these two factors is even. Also, the remaining factor $3$ can also be cancelld: If neither of $k+1$, $k+2$ is a multiple of $3$, then $k$ itself must be a multiple of $3$, but then the third factor $2k+3$ is also a multiple of $3$.

The only somewhat useful alternative writing for the expression is obtained by multiplying out all factors, getting $\frac{2k^3+9k^2+13k+6}{6} =\frac13k^3+\frac32k^2+\frac{13}{6}k+1$.

Hagen von Eitzen
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$$ \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}=\frac{(k+1)(k+2)(2k+3)}{6} $$

Then if $k=2n$ you have

\begin{align*} \frac{(k+1)(k+2)(2k+3)}{6}=& \frac{(2n+1)(2n+2)(4n+3)}{6}\\ =&\frac{(2n+1)2(n+1)(4n+3)}{6}\\ =&\frac{(2n+1)(n+1)(4n+3)}{3} \end{align*}

it's not so different if $k=2n+1$. Then if you want you can consider in more detail: what if $n=2m$? And $n=2m+1$? Enjoy!

Joe
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