" (i) "^(n)C_(r)=(n-r+1)/(r)*^(n)C_(r-1)

Show that , .^(n)C_(r)=(n-r+1)/(r).^(n)C_(r-1) .

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))

Let n and r be non negative integers such that 1<=r<=n* Then,^(n)C_(r)=(n)/(r)*^(n-1)C_(r-1)

If 1<=r<=n, then n^(n-1)C_(r)=(n-r+1)^(n)C_(r-1)

Verify that ""^(n)C_(r )=(n)/(r ) ""^(n-1)C_(r-1) where n=6 and r=3 .

Show that , (.^(n)C_(r)+^(n)C_(r-1))/(.^(n)C_(r-1)+^(n)C_(r-2))=(.^(n+1)p_(r))/(r.^(n+1)p_(r-1))

Prove that if 1 le r le n " then " n xx^((n-1))C_(r-1)= (n-r+1).^(n)C_(r-1)

If sum_(r=0)^(n){("^(n)C_(r-1))/('^(n)C_(r )+^(n)C_(r-1))}^(3)=(25)/(24) , then n is equal to

Find the sum sum_(r=1)^(n) r^(2) (""^(n)C_(r))/(""^(n)C_(r-1)) .