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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

`730`

B

`725`

C

`729`

D

`750`

Text Solution

Verified by Experts

The correct Answer is:
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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