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Mathematics 16 Online
OpenStudy (anonymous):

Please help Binomial theorem

OpenStudy (anonymous):

With?

OpenStudy (anonymous):

Simplify \[\huge \frac{ \left(\begin{matrix}2n \\ n\end{matrix}\right) }{ \left(\begin{matrix}2n-1 \\ n\end{matrix}\right) }\]

OpenStudy (anonymous):

Yep, I cant help you lol. Sorry Ill tag someone @geerky42 You seem smart do you know?

OpenStudy (anonymous):

no prob

geerky42 (geerky42):

\[\binom{n}{k} = \dfrac{n!}{k!(n-k)!}\] So \(\dfrac{\binom{2n}{n}}{\binom{2n-1}{n}} = \dfrac{\dfrac{(2n)!}{n!(2n-n)!}}{\dfrac{(2n-1)!}{n!(2n-n-1)!}}\) After simplifying a little bit and canceling \(n!\) out, we should have \[\dfrac{(2n)!}{(2n-1)!}\cdot\dfrac{(2n-n-1)!}{(2n-n)!} = \dfrac{(2n)!}{(2n-1)!}\cdot\dfrac{(n-1)!}{n!}\] Can you do the rest? Just remember that \(a! = a(a-1)(a-2)\ldots\), so \((a-1)! = (a-1)(a-2)(a-3)\ldots\)

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