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Question
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
Solution
There are 2n + 1 terms in expansion of (1 + x)2n.
∴ The middle term is tn+1.
`"t"_(n+1) = 2n"C"_"n" (1)^(2n - n) x^n = 2n"C"_nx^n = (|__2n)/(|__n |__n) x^n`
`= ((2n)(2n - 1)(2n - 2)(2n - 3)...5*4*3*2*1)/(|__n|__n) x^n`
`= (((2n - 1)(2n - 3)...3*1)2n(2n - 2)...4*2)/(|__n|__n)`
`= ((2n - 1)(2n - 3) ...1 2^n n(n - 1)1...3*2*1)/(|__n|__n)`
`= ((2n - 1)(2n - 3)...3*1)/(|__n) 2^n x^n`
`= (1 * 3 * 5 * 7...(2n - 1)2^n*x^n)/(|__n)`
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