If \(\genfrac{\langle}{\rangle}{0pt}{}{n}{r}\) is defined as in Exercise 24, show that \[\genfrac{\langle}{\rangle}{0pt}{}{n}{r} = \frac{n}{r}\genfrac{\langle}{\rangle}{0pt}{}{n+1}{r-1} = \frac{n+r-1}{r}\genfrac{\langle}{\rangle}{0pt}{}{n}{r-1}\]

We can show that all three components of this equation can be reduced to the same result, thus are equal.

1. \[\genfrac{\langle}{\rangle}{0pt}{}{n}{r} = \binom{n+r-1}{r} = \frac{(n+r-1)!}{r!(n-1)!}\]
2. \[\frac{n}{r}\genfrac{\langle}{\rangle}{0pt}{}{n+1}{r-1} = \frac{n}{r}\frac{(n+r-1)!}{(r-1)!n!} = \frac{(n+r-1)!}{r!(n-1)!}\]
3. \[\frac{n+r-1}{r}\genfrac{\langle}{\rangle}{0pt}{}{n}{r-1} = \frac{n+r-1}{r}\binom{n+r-2}{r-1} = \frac{n+r-1}{r}\frac{(n+r-2)!}{(r-1)!(n-1)!} = \frac{(n+r-1)!}{r!(n-1)!}\]