The number of real negavitve terms in the binomial
expansion of `(1 + ix)^(4n-2) , n in N , n gt 0 , I = sqrt(-1) , ` is

Options

A. n
B. `n+1`
C. `n-1`
D. 2n

Answer

a

Explanation

` (1 + ix) ^(4n -2) = ""^(4n -2) C_(0) + ""^(4n -2)C_(1) (ix) + ""^(4n-2)C_(2)(ix)^(2) + ...+ ""^(4n- 2)C_(4n -2) (ix) ^(4n -2)`
Here , we get see that Ist negative term is ` T_(3)` and the next term is ` T_(7) ` and the last negative term is ` T_(4n-1)`.
Now , ` 3,7,..., 4n-1`
It is an AP.
`because l = a + (N -1)d`
` therefore 4n-1= 3 + (N-1)4`
` rArr n-1= N-1 rArr N=n`