Question

$\Delta_{r} \begin{vmatrix}r&2r -1 &3r - 2\\ \frac{n}{2}&n -1 &a\\ \frac{1}{2}n\left(n - 1\right)&\left(n - 1\right)^{2}&\frac{1}{2}\left(n - 1\right)\left(3n + 4\right)\end{vmatrix}$
then the value of $\sum\limits^{n - 1}_{r = 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)

$\sum\limits^{n - 1}_{r = 1} \Delta_{r} r=1+2 + 3 + ... + \left(n-1\right)=\frac{n\left(n-1\right)}{2}$
$\sum\limits^{n - 1}_{r = 1} \Delta_{r} \left(2-1\right) =1+3 + 5 + ... + \left[2 \left(n - 1\right) - 2\right] = \left(n- 1\right)^{2}$
$\sum\limits^{n - 1}_{r = 1} \Delta_{r} \left(3r-2\right) =1+4 + 7 + .. + \left(3n-3-2\right)$
$= \frac{\left(n-1\right)\left(3n-4\right)}{2}$
$\therefore \sum\limits^{n - 1}_{r = 1} \Delta_{r} $
$= \begin{vmatrix}\sum r&\sum\left(2r-1\right) &\sum \left(3r-2\right)\\ \frac{n}{2}&n-1&a\\ \frac{n\left(n-1\right)}{2}&\left(n-1\right)^{2}&\frac{\left(n-1\right)\left(3n-4\right)}{2}\end{vmatrix}$
$\sum\limits^{n - 1}_{r = 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\limits^{n - 1}_{r = 1} \Delta_{r} $
$= \begin{vmatrix}\frac{n\left(n-1\right)}{2}&\left(n-1\right)^{2} &\frac{\left(n-1\right)\left(3n-4\right)}{2}\\ \frac{n}{2}&n-1&a\\ \frac{n\left(n-1\right)}{2}&\left(n-1\right)^{2}&\frac{\left(n-1\right)\left(3n-4\right)}{2}\end{vmatrix}$
$= 0$
($\because R_{1}$ and $R_{3}$ are identical)
Hence, value of $\sum\limits^{n - 1}_{r = 1} \Delta_{r} $ is independent of both 'a' and 'n'.