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4nC(2n):^(2n)C(n)=(1*3*5*...*(4n-1))/({1...

4nC_(2n):^(2n)C_(n)=(1*3*5*...*(4n-1))/({1*3*5*...*(2n-1)}^(2))

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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))

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)) .

c_(1)^(2)+2C_(2)^(2)+3C_(3)^(2)+....+nC_(n)^(2)=((2n-1)!)/ ([(n-1)!^(2)))

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

(1^(3)+2^(3)+...+n^(3))/(1+3+5+...+(2n-1))=((n+1)^( 2))/(4)