The number of real negative terms in the binomial expansion of (1+ix)(4n-2),  n  N and x > 0 is?

Joshi sir comment

Total number of terms in the expansion = 4n-2+1 = 4n-1

for a negative real number i2 is compulsory so associated terms are 3rd, 7th, 11th ..........

Total number of terms = n

Read 1 Solution.

Write the general expansion of Tr+1 =4n-2Cr(1)4n-2-r( ix)r here for negative terms i should be 2,6,10............(4n-2).

this is an ap with a=2, d=4  and l=4n-2 .you have to find number of terms using the formula l=a+(n-1)d

there fore number of terms =n ans.{rembember r cannot be 4,8,12,..... as i4 =1}

NIKHIL VARSHNEY 10 year ago is this solution helpfull: 1 0

Submit Your Answer

please login to submit your answer

Login Here