A) 26
B) 27
C) 337
D) 378
Correct Answer: B
Solution :
observe \[{{A}_{1}}{{A}_{2}}\] \[=\left[ \begin{align} & 1\,\,\,1 \\ & 0\,\,1 \\ \end{align} \right]\left[ \begin{align} & 1\,\,2 \\ & 0\,\,1 \\ \end{align} \right]=\left[ \begin{align} & 1\,\,3 \\ & 0\,\,1 \\ \end{align} \right]=\left[ \begin{align} & 1\left( 2+1 \right) \\ & 0\,\,\,\,\,\,\,1 \\ \end{align} \right]\] |
And \[{{A}_{1}}{{A}_{2}}{{A}_{3}}=\left[ \begin{align} & 1\,\,3 \\ & 0\,\,1 \\ \end{align} \right]\left[ \begin{align} & 1\,\,3 \\ & 0\,\,1 \\ \end{align} \right]=\left[ \begin{align} & 1\left( 3+2+1 \right) \\ & 0\,\,\,\,\,\,\,\,\,\,\,\,1 \\ \end{align} \right]\] |
So, in general \[{{A}_{1}}{{A}_{2}}{{A}_{3}}...{{A}_{n}}\]\[=\left[ \begin{align} & 1n+\left( n-1 \right)+\left( n-2 \right)+...+3+2+1 \\ & 01 \\ \end{align} \right]\] |
\[So\frac{n\left( n+1 \right)}{2}=378\Rightarrow n=27\] |
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