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Question

The absolute value of middle term in the expansion of (1−1x)n.(1−x)n is

A
2nCn
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B
2nCn
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C
2nCn1
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D
none of these
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Solution

The correct option is D 2nCn
Given, (11x)n(1x)n
=1xn(x1)n(1x)n
=(1)nxn[1x]2n
Therefore the middle term will be
Tn+1=2nCnxn((1)nxn)
=(1)n2nCn

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