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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` |
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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` |
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