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Question
Find the middle terms in the expansion of
`(x + 1/x)^11`
Solution
General term is tr+1 = nCr xn-r ar
Here x is x, a is 1x and n = 11, which is odd.
So the middle terms are `("t"_(n+1))/2 = ("t"_(11+1))/2, ("t"_(n+3))/2 = ("t"_(11+3))/2`
i.e. the middle terms are t6, t7
Now `"t"_6 = "t"_(5+1) = 11"C"_5 x^(11-5) (1/x)^5 = 11"C"_5 x^6 (1/x^5) = 11"C"_5 x`
`"t"_7 = "t"_(6+1) = 11"C"_6 x^(11-6) (1/x)^6`
`= 11"C"_6x^5 (1/x^6) = 11"C"_6(1/x) = 11"C"_(11-6) (1/x) = 11"C"_5 (1/x)`
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