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

$a$ અને $n$ થી પર આધારિત નથી

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 $'n'$.

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