Question

If f, g, h are polynomials of second degree in n having (n+2)a common factor then prove that $\sum _{r=1}^n{△}_r$ is independent on n where
${△}_r$ = $\left|\begin{array}{ccc}2r+1& 6n(n+2)& f\\ 2^{r-1}& 3\cdot 2^{n+1}-6& g\\ r(n-r+1)& n(n+1)(n+2)& h\end{array}\right|$ =

Answer 1

$3+5+7+..............=\frac{n}{2}[2,3+(n-1\left).2\right]=n(n+2)$
$2^0+2^1+2^2+..........+2^{n-1}=\frac{1.(2^n-1)}{2-1}=2^n-1$
$=\frac{1}{6}({6.2}^n-6)=\frac{1}{6}({3.2}^{n+1}-6)$
$1.n+2(n-1)+3(n-2)+3(n-2)+.....+n(n-1)$
= 1.(n+1-1)+2(n+1-2)+3(n+1-3)+......
= $(n+1)(1+2+3+.........)-(1^2+2^2+3^2+..........)$
=$(n+1)\frac{n(n+1)}{6}-\frac{n(n+1)(2n+1)}{6}$
=$\frac{n(n+1)}{6}[3n+3-(2n+1\left)\right]$
=$\frac{1}{6}n(n+1)(n+2)$
Thus putting r= 1,2,3, .......n. we have $\sum _{r=1}^n{△}_r$ becomes a determinant in which $C_1$ and $C_2$ become identical.
Hence $\sum _{r=0}^n{△}_r=0.$