Hello,

does anyone know why:

[(n+3)/n]^n = [(n+3)/(n+2)]^n * [(n+2)/(n+1)]^n *[(n+1)/n]^n

I try but no success.

Thank you guys!

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- May 31st 2011, 04:16 AMMelsitrying to understand an n power algebraic formula!
Hello,

does anyone know why:

[(n+3)/n]^n = [(n+3)/(n+2)]^n * [(n+2)/(n+1)]^n *[(n+1)/n]^n

I try but no success.

Thank you guys! - May 31st 2011, 04:24 AMProve It
$\displaystyle \displaystyle \begin{align*} \left(\frac{n+3}{n+2}\right)^n \cdot \left(\frac{n+2}{n+1}\right)^n \cdot \left(\frac{n+1}{n}\right)^n &= \frac{(n+3)^n}{(n+2)^n}\cdot\frac{(n+2)^n}{(n+1)^n }\cdot\frac{(n+1)^n}{n^n} \\ &= \frac{(n+3)^n\cdot (n+2)^n\cdot (n+1)^n}{(n+2)^n \cdot (n+1)^n \cdot n^n} \\ &= \frac{(n+3)^n}{n^n} \\ &= \left(\frac{n+3}{n}\right)^n\end{align*}$

- May 31st 2011, 04:25 AMMelsi
You are the man!

Thank you very very much for your help!