Question

$\left|\begin{array}{ccc}1& 1& 1\\ {}^m{C}_1& {}^{m+1}{C}_1& {}^{m+2}{C}_1\\ {}^m{C}_2& {}^{m+1}{C}_2& {}^{m+2}{C}_2\end{array}\right|=1$ Given $mϵNm\ge 2$

Answer 1

$\left|\begin{array}{ccc}1& 1& 1\\ {}^m{C}_1& {}^{m+1}{C}_1& {}^{m+2}{C}_1\\ {}^m{C}_2& {}^{m+1}{C}_2& {}^{m+2}{C}_2\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ m& m+1& m+2\\ \frac{m(m+1)}{2}& \frac{(m+1)m}{2}& \frac{(m+2)(m+1)}{2}\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ m& m+1& m+2\\ m(m-1)& (m+1)\left(m\right)& (m+2)(m+1)\end{array}\right|$
$=\frac{1}{2}(1\left(\right(m+1\left)\right(m+2\left)\right(m+1\left)\right)-1\left(m\right(m+2\left)\right(m+1)-(m+2\left)\right(m\left)\right(m-1\left)\right)+1\left(m\right(m+1\left)\right(m)-(m+1\left)m\right(m-1\left)\right))$
$=\frac{1}{2}\left(\right(m+1\left)\right(m+2\left)\right(m+1-m)-m(m+2\left)\right(m+1-m+1)+m(m+1\left)\right(m-m+1\left)\right)$
$=\frac{1}{2}\left(\right(m+1\left)\right(m+2)-2m(m+2)+m(m+1\left)\right)$
$=\frac{1}{2}\left(2\right)=1$