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!)`.

Answer 1

There are (2n + 1) terms in the expansion.

∴ t_n+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`