If `C_r` stands for `nC_r`, then the sum of the series `(2(n/2)!(n/2)!)/(n !)[C_0^2-2C_1^2+3C_2^2-........+(-1)^n(n+1)C_n^2]` ,where n is an even positive integer, is

If C_(r) stands for ""^(n)C_(r) , then the sum of the series (2((n)/(2))!((n)/(2))!)/(n!)[C_(0)^(2)-2C_(1)^(2)+3C_(2)^(2)-...+(-1)^(n)(n+1)C_(n)^(2)] , where n is an even positive integers, is:

If n ge 2 is a positive integer, then the sum of the series ""^(n+1)C_2 + 2(""^2C_2+ ""3C_2 + ""4C_2 +...+ ""nC_2) is:

Find the sum : 1/2*""^(n)C_0+""^(n)C_1+2*""^(n)C_2+2^2*""^nC_3+...+2^(n-1)*""^(n)C_n .

Show that C_1^2-2C_2^2+3.C_3^2- ……….-2n.C_(2n)^2=(-1)^(n-1).n.C_n where C_r stands for ''^(2n)C_r''

Prove that C_(0)2^(2)C_(1)+3C_(2)4^(2)C_(3)+...+(-1)^(n)(n+1)^(2)C_(n)=0 where C_(r)=nC_(r)