The number of terms which are free from radical signs in the expansion of y1/5+x1/1055, is
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answer is 6.
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Detailed Solution
The general term in the expansion of y1/5+x1/1055s given by Tr+1 =55Cr y1/555−rx1/10r=55C ry11−r/5xr/10clearly, Tr+1 will be independent of radical signs if r/5 andr/10 are integers, where 0≤r≤55.r=0 ,10 ,20 ,30 ,40 ,50Hence, there are 6 `terms in the expansion of the given expression which are independent of radicals.