For all positive integers n, show that ^ 2n C_ n + ^ 2n C_ n-1 =1 2 ( ^ 2n+2 C_ n+1 ).

We have, L.H.S.= ^ 2n C_ n + ^ 2n C_ n-1 = 2n! n!n! + 2n! (n-1)!(n-1)! =(2n)! [1 n(n-1)!(n)(n-1)! +1 (n-1)!(n-1)! ] = (2n!) (n-1)!(n-1)! [ 1+n^ 2 n^ 2 ] ...(i) ^ 2n+2 C_ n+1 = (2n+2)! (n+1!)(n+1)! = (2n+2)(2n+1)(2n!) n(n+1)n(n-1)! ...(ii) = (2n!) (n-1)!(n-1)! (n+1)^ 2 (n^ 2 )(n-1)!(n-1)! (2n-2)(2n+1)(2n!) ( 1+n^ 2 n^ 2 ) = (n+1)(n^ 2 +1) (2n+1) 1 2

For all positive integers n, show that ^ 2n C_ n + ^ 2n C_ n-1 =1…

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