If matrix A = begin{bmatrix} 1 & 3k + frac{1} | vedclass

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(B) Let $A_k = \begin{bmatrix} 1 & 3k + \frac{1}{3} \ 0 & 1 \end{bmatrix}$.We know that $\begin{bmatrix} 1 & a \ 0 & 1 \end{bmatrix} \begin{bmatrix}

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$\begin{bmatrix} 1 & 2010 \ 0 & 1 \end{bmatrix}$

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(B) Let $A_k = \begin{bmatrix} 1 & 3k + \frac{1}{3} \ 0 & 1 \end{bmatrix}$.We know that $\begin{bmatrix} 1 & a \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \ 0 & 1 \end{bmatrix}$.Therefore,the product $\prod_{k=1}^{36} A_k = \begin{bmatrix} 1 & \sum_{k=1}^{36} (3k + \frac{1}{3}) \ 0 & 1 \end{bmatrix}$.Calculate the sum: $\sum_{k=1}^{36} (3k + \frac{1}{3}) = 3 \sum_{k=1}^{36} k + \sum_{k=1}^{36} \frac{1}{3}$.$= 3 	imes \frac{36 	imes 37}{2} + 36 	imes \frac{1}{3}$.$= 3 	imes 18 	imes 37 + 12$.$= 54 	imes 37 + 12 = 1998 + 12 = 2010$.Thus,the product is $\begin{bmatrix} 1 & 2010 \ 0 & 1 \end{bmatrix}$.

If matrix $A = \begin{bmatrix} 1 & 3k + \frac{1}{3} \\ 0 & 1 \end{bmatrix}$,then the value of $\prod_{k=1}^{36} \begin{bmatrix} 1 & 3k + \frac{1}{3} \\ 0 & 1 \end{bmatrix}$ is equal to :-

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