Question

If the $n^{th}$ term of the $r^{th}$ order of figurate numbers is equal to the $(n+2{)}^{th}$ term of the $(r-2{)}^{th}$ order$,$ show that $r=n+2$.

Answer 1

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

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

Where as the ${(n+2)}^{th}$ term of the ${(r-2)}^{th}$ order is

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

$=\frac{(n+r-2)!}{(n+1)!(r-3)!}$

$n(n+1)=(r-1)(r-2)$

$n=r-2$

or $r=n+2$

Hence proved