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

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Question

$\sum\limits_{r = 1}^{n - 1} {r = 1 + 2 + 3 + ... + \left( {n - 1} \right)}  = \frac{{n\left( {n - 1} \right)}}{2}$

$\sum\limits_{r = 1}^{n -

Vedclass · Accepted Answer

is independent of both $a$ and $n$

Vedclass · Answer

$\sum\limits_{r = 1}^{n - 1} {r = 1 + 2 + 3 + ... + \left( {n - 1} ight)}  = \frac{{n\left( {n - 1} ight)}}{2}$

$\sum\limits_{r = 1}^{n - 1} {\left( {2r - 1} ight) = 1 + 3 + 5} $

                            $ + ... + \left[ {2\left( {n - 1} ight) - 2} ight] = {\left( {n - 1} ight)^2}$

$\sum\limits_{r = 1}^{n - 1} {\left( {3r - 2} ight)}  = 1 + 4 + 7 + .. + \left( {3n - 3 - 2} ight)$

                           $ = \frac{{\left( {n - 1} ight)\left( {3n - 4} ight)}}{2}$

$	herefore \sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $

$ = \begin{array}{*{20}{c}}
{\sum r }&{\sum {\left( {2r - 1} ight)} }&{\sum {\left( {3r - 2} ight)} }\
{\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}$

$\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.

$	herefore \sum\limits_{r = 1}^{n - 1} {{\Delta _r}} $

$ = \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}$

($\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 $'n'$.

List - $I$	List - $II$
($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has	($1$) a unique solution
($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has	($2$) no solution
($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has	($3$) infinitely many solutions
($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has	($4$) $x=11, y=-2$ and $z=0$ as a solution
	($5$) $x=-15, y=4$ and $z=0$ as a solution

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