Correct option: D.is independent of both $a$ and $n$
d
$\sum\limits_{r = 1}^{n - 1} {r = 1 + 2 + 3 + ... + \left( {n - 1} \right)} = \frac{{n\left( {n - 1} \right)}}{2}$
$\sum\limits_{r = 1}^{n - 1} {\left( {2r - 1} \right) = 1 + 3 + 5} $
$ + ... + \left[ {2\left( {n - 1} \right) - 2} \right] = {\left( {n - 1} \right)^2}$
$\sum\limits_{r = 1}^{n - 1} {\left( {3r - 2} \right)} = 1 + 4 + 7 + .. + \left( {3n - 3 - 2} \right)$
$ = \frac{{\left( {n - 1} \right)\left( {3n - 4} \right)}}{2}$
$\therefore \sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $
$ = \begin{array}{*{20}{c}}
{\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}$
$\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $ consists of $(n-1)$ determinats in $L.H.S.$ and in $R.H.S.$ every constituents of frist row consists of $(n-1)$ elements and hence it can be splitted into sum of $(n-1)$ determinats.
$\therefore \sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $
$ = \begin{array}{*{20}{c}}
{\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}$
($\because $ ${R_1}$ and ${R_3}$ are identical)
Hence, value of $\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $ is independent of both $'a'$ and $'n'$.