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The number of real negative terms in the...

The number of real negative terms in the binomial expansion of `(1+i x)^(4n-2),n in N ,x >0` is `n` b. `n+1` c. `n-1` d. `2n`

A

n

B

`n+1`

C

`n-1`

D

2n

Text Solution

Verified by Experts

The correct Answer is:
a
` (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`
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