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