`C_0C_2+C_1C_3+C_2C_4+c_3C_5+...+C_(n-2)C_n`

C_0C_2 + C_1C_3 +C_2C_4+……..+C_(n-2) C_n =

C_0C_2 + C_1C_3 +C_2C_4+……..+C_(n-2) C_n =

Prove that C_0.C_3 + C_1.C_4 + C_2.C_5 + …..+C_(n-3).C_n = ""^(2n)C_(n +3)

Prove that C_0.C_3 + C_1.C_4 + C_2.C_5 + …..+C_(n-3).C_n = ""^(2n)C_(n +3)

C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+c_(3)C_(5)+...+C_(n-2)C_(n)

(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+….+C_(n)x^(n) then C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+…..+C_(n-2)C_(n) is equal to :

(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+….+C_(n)x^(n) then C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+…..+C_(n-2)C_(n) is equal to :

If C_(0),C_(1),C_(2)…….,C_(n) are the combinatorial coefficient in the expansion of (1+x)^n, n, ne N , then prove that following C_(1)+2C_(2)+3C_(3)+..+n.C_(n)=n.2^(n-1) C_(0)+2C_(1)+3C_(2)+......+(n+1)C_(n)=(n+2)C_(n)=(n+2)2^(n-1) C_(0),+3C_(1)+5C_(2)+.....+(2n+1)C_n =(n+1)2^n (C_0+C_1)(C_1+C_2)(C_2+C_3)......(C_(n-1)+C_n)=(C_0.C_1.C_2....C_(n-1)(n+1)^n)/(n!) 1.C_0^2+3.C_1^2+....+ (2n+1)C_n^2=((n+1)(2n)!)/(n! n!)

Prove that C_0C_2+C_1C_3+…+C_(n-2)C_n=^(2n)C_(n-2)