Question

If $D_r=$
$\left|\begin{array}{ccc}r& x& \frac{n(n+1)}{2}\\ 2r-1& y& n^2\\ 3r-2& z& \frac{n(3n-1)}{2}\end{array}\right|$
then prove that $\sum _{r=1}^n=D_r=0$

Answer 1

$\sum _{r=1}^{n}r=1+2+3+..........+n=\frac{n(n+1)}{2}$ $\sum _{r=1}^n(2r-1)=1+3+5+............+(2n-1)$ $\frac{n}{2}[1+(2n-1\left)\right]n^2$ $\frac{(}{3}r-2)=1+4+7+...........+(3n-2)$ $\frac{n}{2}[1+3n-2]=\frac{n(3n-1)}{2}$. We have used the formula for A.P. $S_n=\frac{n}{2}[a+l]$ $\therefore \sum _{r=1}^n{D}_r=$  $\left|\begin{array}{ccc}\sum r& ;x& ;\frac{n(n+1)}{2}\\ \sum (2r-1)& ;y& ;n^2\\ \sum (3r-2)& ;z& ;\frac{n(3n-1)}{2}\end{array}\right|$  $\sum _{r=1}^{n}Dr$ consists of n determinants in L.H.S. in each of which elements of 1 st column vary but elements of 2 nd and 3rd column remain the same. Hence adding all such determinants, we have sigma in 1st column and no change in 2nd and 3rd column. Now putting the values of various sigma as given in the beginning, we get. $\sum _{r=1}^{n}Dr=$ $\left|\begin{array}{ccc}\frac{n(n+1)}{2}& ;x& ;\frac{n(n+1)}{2}\\ n^2& ;y& ;n^2\\ \frac{n(3n-1)}{2}& ;z& ;\frac{n(3n-1)}{2}\end{array}\right|$ = 0