If the expansion (x2+1x3)n is to contain an independent term, then what should be the value of n?
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Solution
General term, Tr+1=nCr⋅xn−r⋅yr, for (x+y)n ⟹ general term of (x2+1x3)n is nCr⋅x2n−2r⋅1x3r=nCr⋅x2n−5r For a term to be independent of x, 2n−5r should be equal to zero. r=25n, since r can take only integral values, n has to be a multiple.of 5.