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Question
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
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Solution
Here x is 3x, a is `x^2/2`, n = 8, which is even.
∴ The only one middle term = `("t"_(n+1))/2 = ("t"_(8+1))/2 = "t"_5`
General term tr+1 = nCr xn-r ar
`"t"_5 = "t"_(4+1) = 8"C"_4 (3x)^(8-4) (x^2/2)^4 = 8"C"_4 (3x)^4 (x^2)^4/2^4`
`= 8"C"_4 3^4x^4 x^8/2^4 = 8"C"_4 3^4/2^4 x^12 = 8"C"_4 81/16 x^12`
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