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3 and 4 .Determinants and Matrices
hard
જો ${U_n} = \left| {\,\begin{array}{*{20}{c}}n&1&5\\{{n^2}}&{2N + 1}&{2N + 1}\\{{n^3}}&{3{N^2}}&{3N}\end{array}\,} \right|$ તો $\sum\limits_{n = 1}^N {{U_n},} $ મેળવો.
A
$0$
B
$1$
C
$-1$
D
એકપણ નહી.
Solution
(a) $\sum\limits_{n = 1}^N {{U_n} = } \left| {\,\begin{array}{*{20}{c}}{\frac{{N(N + 1)}}{2}}&1&5\\{\frac{{N(N + 1)(2N + 1)}}{6}}&{2N + 1}&{2N + 1}\\{{{\left\{ {\frac{{N(N + 1)}}{2}} \right\}}^2}}&{3{N^2}}&{3N}\end{array}\,} \right|$
$ = \frac{{N(N + 1)}}{{12}}\left| {\,\begin{array}{*{20}{c}}6&1&5\\{4N + 2}&{2N + 1}&{2N + 1}\\{3N(N + 1)}&{3{N^2}}&{3N}\end{array}\,} \right|$
$ = \left| {\,\begin{array}{*{20}{c}}6&1&6\\{4N + 2}&{2N + 1}&{4N + 2}\\{3N(N + 1)}&{3{N^2}}&{3N(N + 1)}\end{array}\,} \right| = 0$,
$\{$Applying ${C_3} \to {C_3} + {C_2}\} $.
Standard 12
Mathematics
