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In the expansion of (7^(1//3)+11^(1//9))...

In the expansion of `(7^(1//3)+11^(1//9))^(6561)` , there are exactly 730 rational term there are exactly 5831 irrational terms the term which involves greatest binomial coefficients is irrational the term which involves greatest binomial coefficients is rational

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

Text Solution

Verified by Experts

The correct Answer is:
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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