Question

${\Delta }_r\left|\begin{array}{ccc}r& 2r-1& 3r-2\\ \frac{n}{2}& n-1& a\\ \frac{1}{2}n\left(n-1\right)& {\left(n-1\right)}^2& \frac{1}{2}\left(n-1\right)\left(3n+4\right)\end{array}\right|$
then the value of $\sum _{r=1}^{n-1}{\Delta }_r$

• A. depends only on a
• B. depends only on n
• C. depends both on a and n
• D. is independent of both a and n

Answer 1

Answer: (D)

$\sum _{r=1}^{n-1}{\Delta }_{r}r=1+2+3+...+\left(n-1\right)=\frac{n\left(n-1\right)}{2}$
$\sum _{r=1}^{n-1}{\Delta }_r\left(2-1\right)=1+3+5+...+\left[2\left(n-1\right)-2\right]={\left(n-1\right)}^2$
$\sum _{r=1}^{n-1}{\Delta }_r\left(3r-2\right)=1+4+7+..+\left(3n-3-2\right)$
$=\frac{\left(n-1\right)\left(3n-4\right)}{2}$
$\therefore \sum _{r=1}^{n-1}{\Delta }_r$
$=\left|\begin{array}{ccc}\sum r& \sum \left(2r-1\right)& \sum \left(3r-2\right)\\ \frac{n}{2}& n-1& a\\ \frac{n\left(n-1\right)}{2}& {\left(n-1\right)}^2& \frac{\left(n-1\right)\left(3n-4\right)}{2}\end{array}\right|$
$\sum _{r=1}^{n-1}{\Delta }_r$ consists of $\left(n-1\right)$ determinants in L.H.S. and in R.H.S every constituent of first row consists of $\left(n-1\right)$ elements and hence it can be splitted into sum of $\left(n-1\right)$ determinants.
$\therefore \sum _{r=1}^{n-1}{\Delta }_r$
$=\left|\begin{array}{ccc}\frac{n\left(n-1\right)}{2}& {\left(n-1\right)}^2& \frac{\left(n-1\right)\left(3n-4\right)}{2}\\ \frac{n}{2}& n-1& a\\ \frac{n\left(n-1\right)}{2}& {\left(n-1\right)}^2& \frac{\left(n-1\right)\left(3n-4\right)}{2}\end{array}\right|$
$=0$
($\because R_1$ and $R_3$ are identical)
Hence, value of $\sum _{r=1}^{n-1}{\Delta }_r$ is independent of both 'a' and 'n'.