Prove that ` ((2n)!)/(n!) =2^(n) xx{1xx3xx5xx...xx(2n-1)}. `

Explanation

We have
` ((2n)!)/(n!) =({1xx2xx3xx4xx5xx...xx(2n-2)xx(2n-1)xx 2n})/(n!) `
`=({1xx3xx5xx...xx(2n-1)}xx{2xx4xx6xx...xx(2n-2)xx2n})/(n!) `
`=({1xx3xx5xx...xx(2n-1)}xx2^(n)xx{1xx2xx3xx...xxn})/(n!) `
` =({1xx3xx5xx...xx(2n-1)}xx2^(n)xx (n!))/(n!) `
` =2^(n)xx{1xx3xx5xx...xx(2n-1)}.`
Hence, `((2n)!)/(n!)= 2^(n)xx{1xx3xx5xx...xx(2n-1)}. `