Binomial Expansion(6)

a)(2x^2+1/√x)^10
= (2x^2 + x^(-1/2))^10
The r+1 th term
T(r+1) = (10Cr) [(2x^2)^(10-r)] [x^(-1/2)]^r
= (10Cr) [(2x^2)^(10-r)] [x^(-r/2)]
= (10Cr) [2^(10-r)] [x^2(10-r)] [x^(-r/2)]
= (10Cr) [2^(10-r)] [ x ^ (20 - 2r - r/2) ]
= (10Cr) [2^(10-r)] [ x ^ (20 - 5r/2) ]

b)
when 20 - 5r/2 = 0 ====> r = 8 ,
the constant term = the (8+1) = 9 th term is (10C8) [2^(10-8)] = 180

when 20 - 5r/2 = 9 ====> r = 22/5 have no integer solution ,
thus it haven't the term x^9.