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Am I correct in stating that if $$\frac{x^n}{x^{n-1}}=\frac{x^n}{x^n\cdot x^{-1}}=\frac{1}{x^{-1}}=x$$ then $$\begin{align*} \left|\frac{(x^2-5x+2)^n}{2^{n+1}}\cdot\frac{2n}{(x^2-5x+2)^{n-1}}\right|&=\left|\frac{(x^2-5x+2)^n}{(x^2-5x+2)^{n-1}}\cdot\frac{2n}{2^{n+1}}\right|\\\\ &\neq\left|(x^2-5x+2)\right|\frac{1}{2} \end{align*}$$

This would have been true if the second term was: $$\frac{2^n}{2^{n+1}}$$ and not $$\frac{2n}{2^{n+1}}$$

I'm trying to confirm that my textbook has a typo to make sure I am not screwing something up in the math.

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Wouldn't it be simpler to write the numerator as $(x^2-5x+2)^{n-1}(x^2-5x+2)$ and simply cancel directly, than to break up the denominator and then "flip" the factor? –  Arturo Magidin May 2 '12 at 20:07
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I suspect this is part of an estimate involving a sequence/series; if so, then the $2n$ is almost certainly a typo for $2^n$, with the general term being $$\frac{(x^2-5x+2)^n}{2^{n+1}}$$ –  Arturo Magidin May 2 '12 at 20:16
    
Yes, much easier to write $x^n=x^{n-1}\cdot x$ to get your initial formula. –  Thomas Andrews May 2 '12 at 20:31
    
@Gigili: Please do not use the [algebra] tag. It was thoroughly discussed on meta, and it was decided to avoid using it. It is still here because we cannot effectively retag all the questions and we cannot really delete it either. So just for future reference, try to avoid adding it to questions. –  Asaf Karagila May 3 '12 at 7:03
    
@AsafKaragila: I think edits overlapped there or something weird happened, I recall I didn't add the $(algebra)$ tag, I don't know why it says I did. I was unaware of the fact, thank you for pointing that out. –  Gigili May 3 '12 at 7:29
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up vote 1 down vote accepted

With the structure of the problem as you've given it, if the supposed answer is $\left|(x^2-5x+2)\right|\frac{1}{2}$, I'd agree that the $2n$ should have been $2^n$.

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