^(4n)C_(2n):^(2n)C_(n)=(1.3.5...(4n-1))/({1.3.5......(2n-1)}^(2))

Prove that : ^(2n)C_n = (2^n [1.3.5. ..........(2n-1)])/(n!) .

Prove that , .^(2n)C_(n)=2^(n)(1.3.5...(2n-1))/(lfloorn)

Prove that: :2^(n)C_(n)=(2^(n)[1.3.5(2n-1)])/(n!)

Find lim_(n rarr oo)((1.3.5...(2n-1)}(n+1)^(4)]+[n^(4)(1.3.5...(2n-1)) (2n+1)]

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)

(1^(4))/(1.3)+(2^(4))/(3.5)+(3^(4))/(5.7)+......+(n^(4)) /((2n-1)(2n+1))=(n(4n^(2)+6n+5))/(48)+(n)/(16(2n+1))

(.^(n)C_(1))/(2)-(2(.^(n)C_(2)))/(3)+(3(.^(n)C_(3)))/(4)-....+(-1)^(n+1)(n(.^(n)C_(n)))/(n+1)=