For all positive integers n, show that
Question:

For all positive integers $n$, show that ${ }^{2 n} C_{n}+{ }^{2 n} C_{n-1}=\frac{1}{2}\left({ }^{2 n+2} C_{n+1}\right)$.

Solution:

$\mathrm{LHS}={ }^{2 n} C_{n}+{ }^{2 n} C_{n-1}$

$=\frac{(2 n) !}{n ! n !}+\frac{(2 n) !}{(n-1) !(2 n-n+1) !}$

$=\frac{(2 n) !}{n ! n !}+\frac{(2 n) !}{(n-1) !(n+1) !}$

$=\frac{(2 n) !}{n(n-1) ! n !}+\frac{(2 n) !}{(n-1) !(n+1) n !}$

$=\frac{(2 n) !}{n !(n-1) !}\left[\frac{1}{n}+\frac{1}{n+1}\right]$

$=\frac{(2 n) !}{n !(n-1) !}\left[\frac{2 n+1}{n(n+1)}\right]$

$=\frac{(2 n+1) !}{n !(n+1) !}$

$\mathrm{RHS}=\frac{1}{2}{ }^{2 n+2} C_{n+1}$

$=\frac{1}{2}\left[\frac{(2 n+2) !}{(n+1) !(2 n+2-n-1) !}\right]$

$=\frac{1}{2}\left[\frac{(2 n+2) !}{(n+1) !(n+1) !}\right]$

$=\frac{1}{2}\left[\frac{(2 n+2)(2 n+1) !}{(n+1) n !(n+1) !}\right]$

$=\frac{1}{2}\left[\frac{2(n+1)(2 n+1) !}{(n+1) n !(n+1) !}\right]$

$=\frac{(2 n+1) !}{n !(n+1) !}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$