`1^2C_1-2^2C_2+3^2C_3- +(-1)^(n-1)n^2C_n=( 1)(n^2*2^(n+1))/(n+1)( 3)(2^(n+1))/(n-1)`

1^(2)C_(1)-2^(2)C_(2)+3^(2)C_(3)-+(-1)^(n-1)n^(2)C_(n)=(1)(n^(2)*2^(n+1))/(n+1)(3)(2^(n+1))/(n-1)

(n+2)C_(0)(2^(n+1))-(n+1)C_(1)(2^(n))+(n)C_(2)(2^(n-1))- is equal to

e_(0)+(C)/(2)+(C_(2))/(3)++(C_(n))/(n+1)=(2^(n+1)-1)/(n+1)

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

Find .^(n)C_(1)-(1)/(2).^(n)C_(2)+(1)/(3).^(n)C_(3)- . . . +(-1)^(n-1)(1)/(n).^(n)C_(n)

2C_(0)+5C_(1)+8C_(2)++(3n+2)C_(n)=(3n+4)2^(n-1)

If quad (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)=((2n)!)/((n!)^(2)) b.((2n)!)/((n-1)!(n+1)!) c.((2n)!)/((n-2)!(n+2)!) d.none of these