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

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

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

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)

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

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 : ""^(n)C_(0).""^(2n)C_(n)-""^(n)C_(1).""^(2n-2)Cn_(n)+""^(n)C_(2).""^(2n-4)Cn_(n)+......=2^n