Show that the middle term in the expansion of `(1+x)^(2n)i s((1. 3. 5 (2n-1)))/(n !)2^n x^n ,w h e r en` is a positive integer.

Explanation

Clearly, the number of terms in the expansion of `(1+x)^(2n)" is "(2n+1).`
`:. " middle term " =((2n)/2+1)" th term "= (n+1)" th term "= T _(n+1).`
Now, `T_(n+1) = .^(2n) C _(n)* x^(n) = ((2n)!)/((n!) xx (n !))* x ^(n) `
`=(1* 2* 3 * 4 ... ( 2n - 2) (2n-1) (2n))/((n!) xx ( n !)) * x^(n)`
`=( [ 1 * 3 * 5 ... ( 2n - 3) ( 2n-1)] xx [2* 4* 6 ... ( 2n - 2) (2n) ])/((n!) (n!)) * x^(n) `
`=([1 * 3* 5 ... ( 2n-1)] xx 2 ^(n) xx [ 1* 2 * 3 ... ( n-1) *n])/((n!) xx (n!))* x^(n)`
`= ([ 1* 3* 5 ... ( 2n-1)] xx 2^(n) xx x^(n))/((n!))*`
Hence , the middle term in the given expansion is
`(1* 3* 5 ... ( 2n-1) xx 2^(n) xx x^(n))/((n!))*`