`.^(2n+1)C_(0)-^((2n+1))C_(1)+^((2n+1))C_(2)-......+(2n+1)C_(2n)=`

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

""^((2n + 1))C_0 + ""^((2n+ 1))C_1 + ""^((2n + 1))C_2 + ……+""^((2n + 1))C_n =

The coefficient of x^(n) in the polynomial (x+""^(2n+1)C_(0))(X+""^(2n+1)C_(1)) (x+""^(2n+1)C_(2))……(X+""^(2n+1)C_(n)) is

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 the equality 1^(2)+2^(2)+3^(2) . . .+n^(2)=.^(n+1)C_(2)+2(.^(n)C_(2)+.^(n-1)C_(2) . . .+.^(2)C_(2)) .

If "^(n)C_(0)-^(n)C_(1)+^(n)C_(2)-^(n)C_(3)+...+(-1)^(r )*^(n)C_(r )=28 , then n is equal to ……

Prove that (^(2n)C_0)^2-(^(2n)C_1)^2+(^(2n)C_2)^2-+(^(2n)C_(2n))^2-(-1)^n^(2n)C_ndot

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))/(.^(n)C_(0))+(2(.^(n)C_(2)))/(.^(n)C_(1))+(3(.^(n)C_(3)))/(.^(n)C_(2))+. . . +(n(.^(n)C_(n)))/(.^(n)C_(n-1))= Sum of first n natural numbers.

Prove that .^(n)C_(1) + 2 xx .^(n)C_(2) + 3 xx .^(n)C_(3) + "…." + n xx .^(n)C_(n) = n2^(n-1) . Hence, prove that .^(n)C_(1).(.^(n)C_(2))^(2).(.^(n)C_(3))^(3)"......."(.^(n)C_(n))^(n) le ((2^(n))/(n+1))^(.^(n+1)C_(2)) AA n in N .

(.^(n)C_(0))^(2)+(.^(n)C_(1))^(2)+(.^(n)C_(2))^(2)+ . . .+(.^(n)C_(n))^(2) equals