1.

In the expansion of `(7^(1/3)+11^(1/9))^(6561)`, the number of terms free from radicals is:A. `730`B. `725`C. `729`D. `750`

Answer» Correct Answer - A
`T_(r+1) =^(6561) C_(r),(7)^((6561-r)/(3))(11)^((r)/(9))`
`rArr r` is multiple of `9`
`:. r = 0,9,18"…….",6561`


Discussion

No Comment Found

Related InterviewSolutions