1.)The number of real negative terms in the binomial expansion of (1+ix) ^4n-2, n belongs to N and x>0 is???Please give the solution with detailed explanation.

Thank You!!

Hi Chaitanya,

i^r will be negative whenever r = 2,6,10....

So all the terms where power of ix is 2,6,10,.....4n-2 will be negative.

So there will be "n" terms that are negative.

Hope it helps.

Wish you all the best.

Regards,

Ashwin (IIT Madaras).

The binomial expansion of (1+ix)^4n-2 is

= ^4n-2C₀ (i)^0 +^4n-2C₁(i)^1 +^4n-2C₂(i)^2+^4n-2C₃(i)^3+^4n-2C₄(i)^4+.....+^4n-2C_4n-1(i)^4n-1+^4n-2C_4n-2(i)^4n-2

Term is real when exponent of i is divisible by 2 and negative when it not divisible by 4. Then exponent  is 2,6,10,...,4n-2

Therefore the no of such terms is n