For a matrix A=12r-101, the value of ∏r=15012r-101 is equal to

A=12r−101=110113011501…………19901=11×3+1×1001501=11×5+1×3+1×10017011901………19901Similarly,=11×1+1×3+1×5+………..+1×9901=1∑r=150 (2r−1)01=1250001

For a matrix A=12r-101, the value of ∏r=15012r-101 is equal to

A=12r−101=110113011501…………19901=11×3+1×1001501=11×5+1×3+1×10017011901………19901Similarly,=11×1+1×3+1×5+………..+1×9901=1∑r=150 (2r−1)01=1250001

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For a matrix $A = [\begin{matrix} 1 & 2 r - 1 \\ 0 & 1 \end{matrix}]$ , the value of $\prod_{r = 1}^{50} [\begin{matrix} 1 & 2 r - 1 \\ 0 & 1 \end{matrix}]$ is equal to

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$[\begin{matrix} 1 & 100 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & 2500 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & 4950 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & 5050 \\ 0 & 1 \end{matrix}]$

answer is D.

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Detailed Solution

$A = [\begin{array}{cc} 1 & 2 r - 1 \\ 0 & 1 \end{array}]$

$\begin{array}{l} = [\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}] [\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}] [\begin{array}{ll} 1 & 5 \\ 0 & 1 \end{array}] \dots \dots \dots \dots [\begin{array}{ll} 1 & 99 \\ 0 & 1 \end{array}] \\ = [\begin{array}{ll} 1 & 1 \times 3 + 1 \times 1 \\ 0 & 0 \end{array}] [\begin{array}{ll} 1 & 5 \\ 0 & 1 \end{array}] \\ = [\begin{array}{cc} 1 & 1 \times 5 + 1 \times 3 + 1 \times 1 \\ 0 & 0 \end{array}] [\begin{array}{ll} 1 & 7 \\ 0 & 1 \end{array}] [\begin{array}{ll} 1 & 9 \\ 0 & 1 \end{array}] \dots \dots \dots [\begin{array}{cc} 1 & 99 \\ 0 & 1 \end{array}] \end{array}$

Similarly,

$\begin{array}{l} = [\begin{array}{cc} 1 & 1 \times 1 + 1 \times 3 + 1 \times 5 + \dots \dots \dots . . + 1 \times 99 \\ 0 & 1 \end{array}] \\ = [\begin{array}{cc} 1 & \sum_{r = 1}^{50} (2 r - 1) \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & 2500 \\ 0 & 1 \end{array}] \end{array}$