Question

Let $f\left(n\right),g\left(n\right),h\left(n\right)$ be polynomials in $n$, and let

${\Delta }_r=\left|\begin{array}{ccc}2r+1& 6n(n+2)& f\left(n\right)\\ 2^{r-1}& {3.2}^{n+1}-6& g\left(n\right)\\ r(n-r+1)& n(n+1)(n+2)& h\left(n\right)\end{array}\right|$

Find the value of $\sum _{r=1}^n{\Delta }_r$.

Answer 1

$\sum _{r=1}^n{\Delta }_r=\left|\begin{array}{ccc}{\sum }_{r=1}^n(2r+1)& 6n(n+2)& f\left(n\right)\\ {\sum }_{r=1}^n{2}^{r-1}& {3.2}^{n+1}-6& g\left(n\right)\\ {\sum }_{r=1}^{n}r(n-r+1)& n(n+1)(n+2)& h\left(n\right)\end{array}\right|$

But

$\sum _{r=1}^n(2r+1)=2\sum _{r=1}^{n}r+n=2\cdot n\frac{(n+1)}{2}+n=n(n+2)$

$\sum _{r=1}^n{2}^{r-1}=2^n-1$

$\sum _{r=1}^{n}r(n-r+1)=(n+1)\sum _{r=1}^{n}r-\sum _{r=1}^n{r}^2$

$=\frac{(n+1)n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=\frac{1}{6}n(n+1)(n+2)$

Thus,

$\sum _{r=1}^n{\Delta }_r=\left|\begin{array}{ccc}n(n+2)& 6n(n+2)& f\left(n\right)\\ 2^n-1& 6(2^n-1)& g\left(n\right)\\ \frac{1}{6}n(n+1)(n+2)& n(n+1)(n+2)& h\left(n\right)\end{array}\right|=0$

$\because C_1$ and $C_2$ are proportional.