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यदि $\Delta_{ r }=\left|\begin{array}{ccc} r & 2 r -1 & 3 r -2 \\ \frac{ n }{2} & n -1 & a \\ \frac{1}{2} n ( n -1) & ( n -1)^{2} & \frac{1}{2}( n -1)(3 n +4)\end{array}\right|$ हैं, तो $\sum_{ r =1}^{ n -1} \Delta_{ r }$ का मान
केवल $a$ पर निर्भर है
केवल $n$ पर निर्भर है
$a$ तथा $n$ दोनों पर निर्भर हैं
$a$ तथा $n$ दोनों से स्वतंत्र हैं।
Solution
$\sum\limits_{r = 1}^{n – 1} {r = 1 + 2 + 3 + … + \left( {n – 1} \right)} = \frac{{n\left( {n – 1} \right)}}{2}$
$\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 $'n'$.
