सिद्ध कीजिए कि - `.^(4n)C_(2n)div^(2n)C_(n)=(1.3.5...(4n-1))/({1.3.5...(2n-1)}^(2))`.

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

Prove that ((4n)C_(2n))/((2n)C_(n))=(1.3.5...(4n-1))/([1.3.5...(2n-1)]^(2))

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

Prove that: (i) r.^(n)C_(r) =(n-r+1).^(n)C_(r-1) (ii) n.^(n-1)C_(r-1) = (n-r+1) .^(n)C_(r-1) (iii) .^(n)C_(r)+ 2.^(n)C_(r-1) +^(n)C_(r-2) =^(n+2)C_(r) (iv) .^(4n)C_(2n): .^(2n)C_(n) = (1.3.5...(4n-1))/({1.3.5..(2n-1)}^(2))

"Prove that "(""^(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))/(lfloorn)

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

Prove that .^(2n)C_(n)=(2^(n)xx[1*3*5...(2n-1)])/(n !) .