Prove that `C_(1)^(2)-2*C_(2)^(2)+3*C_(3)^(2)-…-2n*C_(2n)^(2)=(-1)^(n)n*C_(n)`

Given that C_(1)+2C_(2)x+3C_(3)x^(2)+...+2nC_(2n)x^(2n-1)=2n(1+x)^(2n-1),whereC_(r)=(2n)!/[r!(2n-r)!];r=0,1,2 then prove that C_(1)^(2)-2C_(2)^(2)+3C_(3)^(2)-...-2nC_(2n)^(2)=(-1)^(n)nC_(n).

If (1+x)^n=underset(r=0)overset(n)C_(r)x^r then prove that C_(1)^2+2.C_(2)^(2)+3.C_(3)^2 +…….+n.C_(n)^(2)=((2n-1)!/((n-1)!)^2

Prove that C_(0)^(2)+C_(1)^(2)+...C_(n)^(2)=(2n!)/(n!n!)

c_(1)^(2)+2C_(2)^(2)+3C_(3)^(2)+....+nC_(n)^(2)=((2n-1)!)/ ([(n-1)!^(2)))

If (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + …+ C_(n) x^(n) , prove that C_(0)^(2) - C_(1)^(2) + C_(2)^(2) -…+ (-1)^(n) *C_(n)^(2)= 0 or (-1)^(n//2) * (n!)/((n//2)! (n//2)!) , according as n is odd or even Also , evaluate C_(0)^(2) + C_(1)^(2) + C_(2)^(2) - ...+ (-1)^(n) *C_(n)^(2) for n = 10 and n= 11 .

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

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

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

Prove that (""^(2n)C_(0))^(2)-(""^(2n)C_(1))^(2)+(""^(2n)C_(2))^(2)-…+(""^(2n)C_(2n))^(2)=(-1)^(n)*""^(2n)C_(n) .