Sum of binomial coefficients $\sum_{k=0}^{n}\binom{k+m}{m}$

Question

Let $n,m \in \mathbb{N}$
Evaluate: $$\sum_{k=0}^{n}\binom{k+m}{m}$$ I don't know what to do, because none of the identities I know match this sum. Help is greatly appreciated.

Answer 1

Consider the recurrence identity $$ \binom{n+1}{m+1} - \binom{n}{m+1} = \binom{n}{m} $$ for $n=k+m$. We get, by telescoping: $$ \begin{eqnarray} \sum_{k=0}^n \binom{k+m}{m} &=& \sum_{k=0}^n \left( \binom{m+k+1}{m+1} - \binom{m+k}{m+1} \right) = \sum_{k=0}^n \left(F(k+1)-F(k) \right) \\ &=& F(n+1)-F(0) = \binom{m+n+1}{m+1} - \binom{m}{m+1} \\ &=& \binom{m+n+1}{m+1} \end{eqnarray}$$

Answer 2

If you experiment with $m=0,1,2$, and $3$, trying the first few values of $n$ for each, you may be able to conjecture the identity, which is then very easy to prove by induction. Your summations are column sums of the matrix of binomial coefficients shown below:

$$\begin{array}{r|cc} n\backslash k&0&1&2&3&4\\ \hline 0&1\\ 1&1&1\\ 2&1&2&1\\ 3&1&3&3&1\\ 4&1&4&6&4&1\\ 5&1&5&10&10&5 \end{array}$$

E.g., with $m=2$ and $n=3$ you have

$$\sum_{k=0}^3\binom{k+2}2=\binom22+\binom32+\binom42=1+3+6=10\;.$$

HINT: The sum is a single binomial coefficient expressible in terms of $m$ and $n$.

Alternative HINT: You can try for a combinatorial explanation of the sum. $\binom{k+m}m$ is the number of ways to choose $m$ numbers from the set $\{1,2,\dots,k+m\}$. It’s also the number of ways to choose $m+1$ numbers from the set $\{0,1,2,\dots,n+m\}$ with the stipulation that $k+m$ is the largest number chosen.

Answer 3

I have a better answer. I made use of the rules that ${m+n} \choose {m}$ = sum all k (C(m,k)*C(n,m-k)), Sum k=0 to n (C(k,m)) = C(n+1,m+1) to begin with since i started the equality backwards C(n+m+1,m+1) = sum k=0 to n (C(m+k,k)) .

It took me more than an hour to solve, and tears of joy I solved it so always try out for yourself.

I hope you can se the picture.

Ciao

Picture