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C(0)-C(1)+C(2)-C(3)+ . . .+(-1)^(n)C(n) ...

`C_(0)-C_(1)+C_(2)-C_(3)+ . . .+(-1)^(n)C_(n)` is equal to

A

`2^(n)`

B

`2^(n)-1`

C

`0`

D

`2^(n-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
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