Question

Show that the $n^{th}$ term of the $r^{th}$ order ot figurate numbers is equals to the $r^{th}$ term of the $n^{th}$ order.

Answer 1

$n^{th}$ term of the $r^{th}$ order is

$\frac{(n+r-2)!}{(n-1)!(r-1)!}$

whereas, $r^{th}$ term of the $n^{th}$ order is

$=\frac{(r+n-2)!}{(r-1)!(n-1)!}$

Hence, they are equal.