91Ó°ÊÓ

Here, we need to prove that $1/B\left(n,m\right)=m\left(\begin{array}{l}n+m-1\\ n-1\end{array}\right)=n\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)$.

The beta function is defined as $\mathit{B}\mathbf{(}\mathit{p}\mathbf{,}\mathit{q}\mathbf{)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{∫}}}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{q}\mathbf{-}\mathbf{1}}\mathit{d}\mathit{x}$$B(p,q)=\underset{0}{\overset{1}{∫}}{x}^{p-1}{(1-x)}^{q-1}dx$.

Write the first fraction in the given equation.

$m\left(\begin{array}{l}n+m-1\\ n-1\end{array}\right)$

Write the relation.

$\left(\begin{array}{l}p\\ n\end{array}\right)=\frac{\mathrm{Γ}(p+1)}{n!\mathrm{Γ}(p-n+1)}$

Continue evaluating the expressions.

$m.\frac{\mathrm{Γ}(n+m-1+1)}{(n-1)\mathrm{Γ}(n+m-1-n+1+1)}=\frac{\mathrm{Γ}(n+m)}{(n-1)\mathrm{Γ}(m+1)}$

Use the identities of evaluating Gamma functions and continue calculations.

$\begin{array}{c}\mathrm{Γ}(p+1)=p\mathrm{Γ}\left(p\right)\\ m·\frac{\mathrm{Γ}(n+m)}{(n-1)!m\mathrm{Γ}\left(m\right)}=\frac{\mathrm{Γ}(n+m)}{(m-1)!\mathrm{Γ}\left(m\right)}\end{array}$

Again, use identities of Gamma functions to further simplify.

$\begin{array}{c}\mathrm{Γ}\left(m\right)=(m-1)!\\ \frac{\mathrm{Γ}(n+m)}{(m-1)!\mathrm{Γ}\left(m\right)}=\frac{\mathrm{Γ}(n+m)}{(n-1)!(m-1)!}\end{array}$

This gives the answer.

Start with the second fraction in the expression.

$n\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)$

Continue evaluating in the same way as before.

$n\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)=\frac{\mathrm{Γ}(n+m)}{(n-1)!(m-1)!}$

Since both sides are equal, the given statement is proved.