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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`, the number of terms free from radicals is:

A

725

B

730

C

731

D

none

Text Solution

Verified by Experts

B
`(7^((1)/(3))+11^((1)/(9)))^(6561)`
General term `=""^(6561)C_(r)(7)^((6561-r)/(3))(11)^((r)/(9))`
For r=0,9,18,27,.........6561.
terms are free from radical sign.
`:.` Total 730 terms.
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