# Prove this product

How to prove this product? $$\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$$

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Are you familiar with mathematical induction? – Austin Mohr Jan 27 '13 at 4:33
Yes, but is that the only way? :) – gauss115 Jan 27 '13 at 4:36
@gauss115 : WHY do you have to type something like {{{k}^{2}}+k+1} where {k^2+k^1} will suffice? This is lunacy. I see it frequently on math.stackexchange.com, and never anywhere else. – Michael Hardy Jan 27 '13 at 4:58
@MichaelHardy Personally I think it's because the $\LaTeX$ formulae are converted by a software, such as MathType. – Frank Science Jan 27 '13 at 5:18
@MichaelHardy Essentially, because things mistakes like $x^-1$, $g_ij$ require recompilation - which is annoying - if you put in plenty of brackets you get more intuitive behaviour. Much in someone who uses supercollider might be tempted to put brackets in ((ab)+(cd)) no matter what language they are using, as they are intuitively aware that syntactic sugar does not apply everywhere. Or it is a result of changing something from x^{-2} to x^{2} - deleting and re-adding curly brackets is very annoying. – Lucas Jan 27 '13 at 14:47

HINT: It never hurts to gather some data by doing some actual computation:

$$\begin{array}{c|l} n&\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}\\ \hline 2&\frac73\\ 3&\frac{\color{red}7}3\cdot\frac{13}{\color{red}7}\\ 4&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{21}{\color{blue}{13}}\\ 5&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{\color{green}{21}}{\color{blue}{13}}\cdot\frac{31}{\color{green}{21}} \end{array}$$

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Hint: If $f(m)=m^2+m+1$ then $f(m-1)=m^2-m+1$. Now there will be a whole lot of cancellation goin' on.

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$$\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}=\frac{\rlap{/}7}{3}\frac{\rlap{\;/}13}{\rlap{/}7}\frac{\rlap{\;/}21}{\rlap{\;/}13}\frac{31}{\rlap{\;/}21}\cdot\ldots$$

Can you see what the only factors that'll remain are?

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