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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!!

290 Points
9 years ago

Hi Chaitanya,

ir 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,

Swapnil Saxena
102 Points
9 years ago

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-2C4n-1(i)^4n-1+4n-2C4n-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