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Question

Let A=111011001. Then for positive integer n, An is

A
1nn20n2n00n
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B
⎢ ⎢ ⎢1nn(n+12)01n001⎥ ⎥ ⎥
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C
1n2n0nn200n2
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D
⎢ ⎢ ⎢ ⎢ ⎢1n2n10n+12n200n+12⎥ ⎥ ⎥ ⎥ ⎥
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Solution

The correct option is B ⎢ ⎢ ⎢1nn(n+12)01n001⎥ ⎥ ⎥
A=111011001

A2=123012001

A3=136013001

Observing the pattern, we can conclude that
An=⎢ ⎢ ⎢1nn(n+1)201n001⎥ ⎥ ⎥

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