`C(n,r)+C(n,r-1)=C(n+1,r)`

Prove that C(n,r)+C(n-1,r)+C(n-2,r)+......+C(r,r)=C(n+1,r+1)

Given that C(n, r) : C(n,r + 1) = 1 : 2 and C(n,r + 1) : C(n,r + 2) = 2 : 3 . What is P(n, r) : C(n, r) equal to ?

If sum_(r=0r=0)^(n){(C(n,r-1))/(C(n,r)+C(n,r-1))}^(3)=(25)/(24)

If sum_(r=1)^(n)r^(3)((C(n,r))/(C(n,r-1)))=14^(2) then n=

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

Let n and r be no negative integers suych that r<=n. Then,^(n)C_(r)+^(n)C_(r-1)=^(n+1)C_(r)

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

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))