If {Delta _r} = left| {begin{array}{*{20}{c}} | vedclass

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(D) We are given the determinant ${\Delta _r} = \left| {\begin{array}{*{20}{c}} r&{2r - 1}&{3r - 2} \ {\frac{n}{2}}&{n - 1}&a \ {\frac{1}{2}n\left(

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is independent of both $a$ and $n$

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(D) We are given the determinant ${\Delta _r} = \left| {\begin{array}{*{20}{c}} r&{2r - 1}&{3r - 2} \ {\frac{n}{2}}&{n - 1}&a \ {\frac{1}{2}n\left( {n - 1} ight)}&{{{\left( {n - 1} ight)}^2}}&{\frac{1}{2}\left( {n - 1} ight)\left( {3n - 4} ight)} \end{array}} ight|$.We need to calculate $\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $. Since the summation operator is linear with respect to the rows of a determinant,we can write:$\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} = \left| {\begin{array}{*{20}{c}} {\sum\limits_{r = 1}^{n - 1} r }&{\sum\limits_{r = 1}^{n - 1} {\left( {2r - 1} ight)} }&{\sum\limits_{r = 1}^{n - 1} {\left( {3r - 2} ight)} } \ {\frac{n}{2}}&{n - 1}&a \ {\frac{1}{2}n\left( {n - 1} ight)}&{{{\left( {n - 1} ight)}^2}}&{\frac{1}{2}\left( {n - 1} ight)\left( {3n - 4} ight)} \end{array}} ight|$.Calculating the sums:$1. \sum\limits_{r = 1}^{n - 1} r = \frac{{\left( {n - 1} ight)n}}{2}$$2. \sum\limits_{r = 1}^{n - 1} {\left( {2r - 1} ight)} = 2\frac{{\left( {n - 1} ight)n}}{2} - \left( {n - 1} ight) = n(n-1) - (n-1) = {(n-1)^2}$$3. \sum\limits_{r = 1}^{n - 1} {\left( {3r - 2} ight)} = 3\frac{{\left( {n - 1} ight)n}}{2} - 2(n-1) = \frac{{3n(n-1) - 4(n-1)}}{2} = \frac{{\left( {n - 1} ight)\left( {3n - 4} ight)}}{2}$Substituting these back into the determinant:$\sum\limits_{r = 1}^{n - 1} {{\Delta _r}} = \left| {\begin{array}{*{20}{c}} {\frac{{n\left( {n - 1} ight)}}{2}}&{{{\left( {n - 1} ight)}^2}}&{\frac{{\left( {n - 1} ight)\left( {3n - 4} ight)}}{2}} \ {\frac{n}{2}}&{n - 1}&a \ {\frac{{n\left( {n - 1} ight)}}{2}}&{{{\left( {n - 1} ight)}^2}}&{\frac{{\left( {n - 1} ight)\left( {3n - 4} ight)}}{2}} \end{array}} ight|$.Since the first row $(R_1)$ and the third row $(R_3)$ are identical,the value of the determinant is $0$.Therefore,the value is independent of both $a$ and $n$.

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