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In the expansion of `(7^(1/3)+ 11^(1/9))^6561`, the number of terms free from radicals is:A. there are exactly `730` rational termsB. there are exactly 5832 irrational termsC. the term which involves greatest binomial coefficients is irrationalD. the term which involves greatest binomial coefficients is rational |
Answer» Correct Answer - A::B::C General term is `.^(6561)C_(r) 7^((6561-r)/(3)) 11^((r)/(9))` To make the term free of radical sign, r should be a multiple of 9. `:. r=0, 9, 18,27,"…..",6561`. Hence, there are 730 rational terms. The greatest binomial coefficients are `.^(6561)C_((6561-1)/(2))` and `.^(6561)C_((6561+11)/(2))` or `.^(6561)C_(3280)` and `.^(6561)C_(3281)`. Now `3280` are `3281` are not a multiple of 3, hence, both terms involving greatest binomial coefficients are irrational . |
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