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:

Prove that sum_(r=1)^(k)(-3)^(r-1)C(3n,2r-1)=0 where k=(3n)/(2) and n is an even positive integer.

The sum of the series 1+(1)/(2) ""^(n) C_1 + (1)/(3) ""^(n) C_(2) + ….+ (1)/(n+1) ""^(n) C_(n) is equal to

The value of hat r(2n+1)C_(0)^(2)+^(2n+1)C_(1)^(2)+^(2n+1)C_(2)^(2)+....+^(2n+1)C_(n)^(2) is equal to

Prove that ^nC_(0)^(n)C_(0)-^(n+1)C_(1)^(n)C_(1)+^(n+2)C_(2)^(n)C_(2)-...=(-1)^(n)

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 .