Question

If a determinant is of the $n^{th}$ order, and if the constituents of its first, second, third, $...n^{th}$ rows are the first $n$ figurate numbers of the first, second, third, $...n^{th}$ orders, show that its value is unity,.

Answer 1

The determinant in question is;-$\left|\begin{array}{ccccccc}1& 1& 1& 1& 1& 1& ...\\ 1& 2& 3& 4& 5& 6& ...\\ 1& 3& 6& 10& 15& 21& ...\\ 1& 4& 10& 20& 35& 56& ...\\ 1& 5& 15& 35& 70& 126& ...\\ ..& ..& ..& ..& ..& ..& ..\\ ..& ..& ..& ..& ..& ..& ..\end{array}\right|$

If a new determinant is formed by subtracting each row from the row immediately beneath it, we obtain a determinant in which each constituent of the first column vanishes except the first; thus the determinant;

$\left|\begin{array}{cccccc}1& 2& 3& 4& 5& ...\\ 1& 3& 6& 10& 15& ...\\ 1& 4& 10& 20& 35& ...\\ 1& 5& 15& 35& 70& ...\\ ..& ..& ..& ..& ..& ..\\ ..& ..& ..& ..& ..& ..\end{array}\right|$

This determinant consists of ${}^{\prime }n-1^{\prime }$ rows and constituents of the successive rows are easily seen to the first $n-1$ terms of the figurate numbers of the second,third,fourth,....$nth$ orders

In like manner the last determinant

$=\left|\begin{array}{ccccc}1& 3& 6& 10& ...\\ 1& 4& 10& 20& ...\\ 1& 5& 15& 35& ...\\ ..& ..& ..& ..& ..\\ ..& ..& ..& ..& ..\end{array}\right|$

The constituents of the successive rows of this determinant are the first $n-1$ terms of the figurate numbers of the third,fourth,...$nth$ orders;-

The determinant at length will reduce to

$\left|\begin{array}{cc}1& n-1\\ 1& n\end{array}\right|$

and therefore its value is unity