For n ∈ {0} ∪N, and k ∈ {0} ∪N, if k ≤ n, deﬁne C(n,k) = n!(k!)−1([n − k]!)−1.

For n ∈ {0} ∪N, and k ∈ {0} ∪N, if k n, deﬁne C(n,k) = 0. Prove that for any m,n ∈ {0} ∪N

m+n

sum

k=m

C(k,m) = C(m + n + 1,m + 1).

Question

For n ∈ {0} ∪N, and k ∈ {0} ∪N, if k ≤ n, deﬁne C(n,k) = n!(k!)−1([n − k]!)−1.

For n ∈ {0} ∪N, and k ∈ {0} ∪N, if k > n, deﬁne C(n,k) = 0. Prove that for any m,n ∈ {0} ∪N

m+n

sum

k=m

C(k,m) = C(m + n + 1,m + 1).

geerky42 · Answer

$\displaystyle \sum_{k=m}^{m+n}C(k,m)=C(m + n + 1,m + 1)$ ?