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જો શ્રેણિક $A\, = \,\left[ {\begin{array}{*{20}{c}}
1&{3k + \frac{1}{3}} \\
0&1
\end{array}} \right]$,તો $\mathop \Pi \limits_{k = 1}^{36} \,\left[ {\begin{array}{*{20}{c}}
1&{3k + \frac{1}{3}} \\
0&1
\end{array}} \right]$ ની કિમંત મેળવો.
$\left[ {\begin{array}{*{20}{c}} 1&{1998} \\ 0&1 \end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}} 1&{2010} \\ 0&1 \end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}} 1&{1005} \\ 0&1 \end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}} 1&{999} \\ 0&1 \end{array}} \right]$
Solution
$\prod\limits_{k = 1}^{36} {\left[ {\begin{array}{*{20}{c}}
1&{3k + \frac{1}{3}}\\
0&1
\end{array}} \right]} $
$ = \left[ {\begin{array}{*{20}{c}}
1&{3 + \frac{1}{3}}\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{6 + \frac{1}{3}}\\
0&1
\end{array}} \right]…….\left[ {\begin{array}{*{20}{c}}
1&{108 + \frac{1}{3}}\\
0&1
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
1&{\left( {3 + 6 + 9 + …. + 108} \right) + 12}\\
0&1
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
1&{2010}\\
0&1
\end{array}} \right]$
