`((2n+2)!)/((2n)!)`

If n is a positive integer,prove that 1-2n+(2n(2n-1))/(2!)-(2n(2n-1)(2n-2))/(3!)+...+(-1)^(n-1)(2n(2n-1)...(n+2))/((n-1)!)=(-1)^(n+1)(2n)!/2(n!)^(2)

If U_(n)=(1+(1)/(n^(2)))(1+(2^(2))/(n^(2)))^(2).............(1+(n^(2))/(n^(2)))^(n) m then lim_(n to oo)(U_(n))^((-4)/(n^(2))) is equal to

In a game a coin is tossed 2n+m xx and a a player wins if he does not get any two consecutive outcomes same for at least 2n xx in a row.The probability that player wins the game is (m+2)/(2^(2n)+1) b.(2n+2)/(2^(2n)) c.(2n+2)/(2^(2n+1)) d.(m+2)/(2^(2n))

Find sum of products,taken two at a time,of coefficients in expansion of (1+x)^(n)[2^(2n-1)-((2n!)/(2(n!)^(2)))]

Simplify: (2^(n+4)-2(2^(n)))/(2(2^(n+3)))

Find the value of ( 2^(n+4) - 2 ( 2^n )) / ( 2(2^(n+3) )

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