^(n)C_(r+1)+2''C_(r)+''C_(r-1)=

Prove that .^(n+1)C_(r+1)+^nC_r+^nC_(r-1)=^(n+2)C_(r+1)

""^(n) C_(r+1)+2""^(n)C_(r) +""^(n)C_(r-1)=

""^(n) C_(r+1)+2""^(n)C_(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))

If ""^(n)C_(r ): ""^(n)C_(r+1)=1:2 and ""^(n)C_(r+1): ""^(n)C_(r+2)=2:3 , find n and r.

Prove that "^nC_r+2 ^(n)C_(r-1)+ ^(n)C_(r-2) = ^(n+2)C_r .

If f(x)=sum_(r=1)^(n) { r^(2) (""^(n)C_(r)- ^(n) C_(r-1))+ (2r+1) ^(n) C_(r)} and f(30)=30(2)^(lambda), then the value of lambda is

If f(x)=sum_(r=1)^(n) { r^(2) (""^(n)C_(r)- ^(n) C_(r-1))+ (2r+1) ^(n) C_(r)} and f(30)=30(2)^(lambda), then the value of lambda is

show that ^nC_r+ ^(n-1)C_(r-1)+ ^(n-1)C_(r-2)= ^(n+1)C_r

Prove that ""^(n)C_(r )+2""^(n)C_(r-1)+ ""^(n)C_(r-2)= ""^(n+2)C_(r ) .