Question

If m and p are positive $(m\ge p)$ and $\Delta (m,p)=\left|\begin{array}{ccc}{}^m{C}_p& {}^m{C}_{p+1}& {}^m{C}_{p+2}\\ {}^{m+1}{C}_p& {}^{m+1}{C}_{p+1}& {}^{m+1}{C}_{p+2}\\ {}^{m+2}{C}_p& {}^{m+2}{C}_{p+1}& {}^{m+2}{C}_{p+2}\end{array}\right|$ and ${}^m{C}_p=0$ if m < p, then

• A. $\Delta (2,1)/\Delta (1,0)=4$
• B. $\Delta (4,3)/\Delta (3,2)=2$
• C. $\Delta (4,3)/\Delta (2,1)=5$
• D. $\Delta (4,3)/\Delta (1,0)=10$

Answer 1

Answer: (A), (B), (C)

$\Delta (2,1)/\Delta (1,0)=4$
$\Delta (4,3)/\Delta (3,2)=2$
$\Delta (4,3)/\Delta (2,1)=5$

$\Delta (m,m-1)=\left|\begin{array}{ccc}{}^m{C}_{m-1}& {}^m{C}_m& 0\\ {}^{m+1}{C}_{m-1}& {}^{m+1}{C}_m& {}^{m+1}{C}_{m+1}\\ {}^{m+2}{C}_{m-1}& {}^{m+2}{C}_m& {}^{m+2}{C}_{m+1}\end{array}\right|$

$=\left|\begin{array}{ccc}m& 1& 0\\ \frac{(m+1)m}{2}& (m+1)& 1\\ \frac{(m+2)(m+1)m}{6}& \frac{(m+2)(m+1)}{2}& (m+2)\end{array}\right|$

$=m(m+2)\left|\begin{array}{ccc}1& 1& 0\\ \left[\frac{m+1}{2}\right]& (m+1)& 1\\ \frac{(m+1)}{6}& \frac{(m+1)}{2}& 1\end{array}\right|$

$=m(m+2)\left|\begin{array}{ccc}1& 1& 0\\ \frac{(m+1)}{3}& \left[\frac{m+1}{2}\right]& 0\\ \left[\frac{m+1}{6}\right]& \frac{(m+1)}{2}& 1\end{array}\right|=\frac{m(m+1)(m+2)}{6}$