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(1+x)^(n)=C(0)+C(1)x+C(2)x^(2)+….+C(n)x^...

`(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+….+C_(n)x^(n)` then `C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+…..+C_(n-2)C_(n)` is equal to :

A

`.^(2n)C_(n-1)`

B

`.^(n)C_(n-1)`

C

`.^(2n)C_(n+1)`

D

`.^(n)C_(n+1)`

Text Solution

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The correct Answer is:
A
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