Show that the coefficient of the middle term in the expansion of `(1+x)^(2n)`is equal to the sum of the coefficients of two middle terms in the expansion of `(1+x)^(2n-1)`.

Explanation

The expansion of `(1+x)^(2n) " contains " (2n +1) " terms and therefore, the " (n+1) ` th term is the middle term.
`:.` the middle term in the expansion of `(1+x)^(2n)` is given by
`t_(n+1) = .^(2n)C_(n)x^(n) = ((2n)!)/((n!)*(n!)) x^(n) = ((2n)!)/((n!)^(2)) x^(n).`
Thus, the coefficient of the middle term of `(1+x)^(2n) " is "((2n)!)/((n!)^(2)).`
Again , the expansion of `(1+x)^(2n-1)` contains 2n terms.
So, the middle terms are the nth and the `(n+1)` th terms.
Now, `t_(n) = t _((n-1)+1)= .^(2n-1) C_(n-1) x^(n-1) and t_(n+1)= .^(2n-1) C _(n)x^(n).`
` :.` the sum of the coefficients of the middle terms of `(1+x)^(2n-1)`
`=.^(2n-1)C _(n-1) + .^(2n-1) C _(n)`
`=((2n-1)!)/((n-1)! xx (n!)) + ((2n-1)!)/((n!) xx (n-1)!)`
` = ((2 xx(2n-1)! xx n))/((n!) xx n xx (n-1)!)" "` [multiplying num. and denom. by n]
`((2n)* ( 2n-1)!)/((n!)*(n!))=((2n)!)/((n!)^(2))`
= the coefficent of the middle term of `(1+x)^(2n)`.