`C_0^2+3*C_1^2+5*C_2^2+.........+(2n+1)*C_n^2=`

Show that C_0 + 3C_1 + 5C_2 + .... +(2n+1) C_n = (n+1)(2^n)

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!)

If n=5 ,then ("^n C_0)^(2) + ("^n C_1)^(2) + ("^n C_2)^(2) +......+ ("^n C_5)^(2) is equal to

C_ (0) ^ (2) + 2C_ (1) ^ (2) + 3.C_ (2) ^ (2) + ............ + (n + 1) C_ (n ) ^ (2) =

prove that :C_(0)^(2)+3C_(1)^(@)+5C_(2)^(2)+...+(2n+1)C_(n)^(2)=((n+1)2n!)/((n!)^(2))