Saved Bookmarks
| 1. |
In the expansion of `(7^(1/3)+ 11^(1/9))^6561`, the number of terms free from radicals is:A. 725B. 730C. 731D. none |
|
Answer» 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. |
|