Show that the middle term in the expansion of `(1+x)^(2n)` is `(1.3.5….(2n-1))/(n!) 2^(n)x^(n)`, where n is a positive integer.

Explanation

The number of terms in the expansion of `(1 + x)^(2x)` is
2n +1 (odd) , its middle term is (n +1)th term
` therefore ` Required term `= T_(n +1)`
`= ""^(2n)C_(5)x^(n)= (2n!)/(n!n!) x^(n) = ((1*2*3*4...(2n -1).2n))/(n!n!)x^(n)`
` = ({1*3*5...(2n-1)}{2*4*6...2n})/(n!n!) x^(n)`
` = ({1*3*5...(2n-1)} 2^(n) (1*2*3...n))/(n!n!) x^(n)`
` = ({ 1*3*5...(2n-1) }2^(n) (1*2*3...n))/(n!n!) x^(n)`

` = ({ 1*3*5...(2n-1) }2^(n) n!) /(n!)x^(n)=( 1*2*3...(2n-1))/(n! )2^(n) x^(n)`