In the expansion of `(7^(1/3)+ 11^(1/9))^6561`, the number of terms free from radicals is:

Options

A. there are exactly `730` rational terms
B. there are exactly 5832 irrational terms
C. the term which involves greatest binomial coefficients is irrational
D. the term which involves greatest binomial coefficients is rational

Answer

A, B, C

Explanation

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 .