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

If `(1+x)^(n) = C_(0) + C_(1)x + C_(2)x^(2) + "….." + C_(n)x^(n)`, then find the value of `C_(0)+C_(2)+C_(4)+C_(6)+........` ​ ​

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The correct Answer is:
B, C
For `n = 2m`, the given expression is
`C_(0)-(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))-(C_(0)+C_(1)+C_(2)+C_(3))+"....."(-1)^(n-1)(C_(0)+C_(1)+"....."+C_(n-1))`
`= C_(0)-(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))-(C_(0)+C_(1)+C_(2)+C_(3))+"....."-(C_(0)+C_(1)+".... "+C_(2m-1))`
`= - (C_(1)+C_(3)+C_(5)+"...."+C_(2m-1))`
`= - (C_(1)+C_(3)+C_(5)+"......."+C_(n-1)) = -2^(n-1)`
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