यदि Delta_{ r }=left|begin{array}{ccc} r | 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

$a$ तथा $n$ दोनों से स्वतंत्र हैं।

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 - $II$
($P$)यदि $\beta=\frac{1}{2}(7 \alpha-3)$ एवं $\gamma=28$, तब निकाय का(के)	($1$) क अद्वितीय हल (unique solution) है
($Q$)यदि $\beta=\frac{1}{2}(7 \alpha-3)$ एवं $\gamma \neq 28$, तब निकाय का(के)	($2$)कोई हल नहीं है
($R$) Iयदि $\beta \neq \frac{1}{2}(7 \alpha-3)$ जहाँ $\alpha=1$ एवं $\gamma \neq 28$, तब निकाय का(के)	($3$)अनंत हल हैं
($S$) यदि $\beta \neq \frac{1}{2}(7 \alpha-3)$ जहाँ $\alpha=1$ एवं $\gamma=28$, तब निकाय का(के)	($4$) $x=11, y=-2$ एवं $z=0$ एक हल है
	($5$) $x=-15, y=4$ एवं $z=0$ एक हल है

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

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