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If `(1+x)^(n) = C_(0) + C_(1)x + C_(2)x^(2) + "….." + C_(n)x^(n)`, then `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))` is (where n is even integer and `C_(r) = .^(n)C_(r)`)A. a positive valueB. a negative valueC. divisible by `2^(n-1)`D. divisible by `2^(n)` |
Answer» Correct Answer - 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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