Find the 13th term in the expansion of (9x−13√x)18,x≠0
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Solution
It is known that(r+1) term (Tr+1) in the binomial expansion of (a+b)n is given by Tr+1=nCran−rbr Thus 13th term in the expansion of (9x−13√x)18 is T13=T12+1=18C12(9x)18−12(−13√x)12 =(−1)1218!12!6!(9)6(x)6(13)12(1√x)12 =18.17.16.15.14.13.12!12!.6.5.4.3.2.x6(1x6).312(1312)