1.

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))

Answer»

a positive value
a negative value
divisible by `2^(n-1)`
divisible by `2^(n)`

SOLUTION :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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