^(2n)C_(2)+^(2n)C_(4)+......+^(2n)C_(2n)=

Evaluate .^(n)C_(0).^(n)C_(2)+.^(n)C_(1).^(n)C_(3)+.^(n)C_(2).^(n)C_(4)+"...."+.^(n)C_(n-2).^(n)C_(n) .

Prove that "^(2n)C_0 + ^(2n)C_2 + .... + ^(2n)C_(2n) = 2^(2n-1)

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

If (1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a+.^(n)C_(2)a^(2)+ . . +.^(n)C_(n)a^(n) , then prove that .^(n)C_(1)+2.^(n)C_(2)+3.^(n)3C_(3)+ . . .+n.^(n)C_(n)=n.2^(n-1) .

If (1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a+.^(n)C_(2)a^(2)+ . . +.^(n)C_(n)a^(n) , then prove that .^(n)C_(1)+2.^(n)C_(2)+3.^(n)3C_(3)+ . . .+n.^(n)C_(n)=n.2^(n-1) .

Prove that "^(2n)C_1 + ^(2n)C_3 + .... + ^(2n)C_(2n-1) = 2^(2n-1)

Prove that (""^(2n)C_(0))^(2)-(""^(2n)C_(1))^(2)+(""^(2n)C_(2))-(""^(2n)C_(3))^(2)+......+(""^(2n)C_(2n))^(2)=(-1)^(n)(""^(2n)C_(n))^2.

Find 'n', if ""^(2n)C_(1), ""^(2n)C_(2) and ""^(2n)C_(3) are in A.P.

If ""^(2n)C_(1), ""^(2n)C_(2) and ""^(2n)C_(3) are in A.P., find n.

STATEMENT - 1 : If n is even, .^(2n)C_(1)+.^(2n)C_(3)+.^(2n)C_(5)+"….."+.^(2n)C_(n-1) = 2^(2n-2) . STATEMENT - 2 : .^(2n)C_(1) + .^(2n)C_(3)+ .^(2n)C_(5) + "……"+ .^(2n)C_(2n-1) = 2^(2n-1)