Sums of binomial coefficients in which lower index remains fixed How can I prove that $ \sum_{k=1}^{m} {{k+n} \choose {n+1}} = {{ m + n + 1 } \choose {n+2}} $?
I have managed to prove that $ \sum_{k=0}^{m} {{k+n} \choose {n}} = {{ m + n + 1 } \choose {n+1}} $
Although it seams intuitive that the second implies the first, I haven't been able to deal with the extra +1.
Thanks in advance
 A: Note that
$$\sum_{k=1}^{m} {{k+n} \choose {n+1}}=\sum_{k=1}^{m} {{k-1+n+1} \choose {n+1}}=\sum_{j=0}^{m-1} {{j+n+1} \choose {n+1}}={{ (m-1) + (n+1) + 1 } \choose {(n+1)+1}}$$
where at the last step we use the identity you have already proved.
A: The given sum $$S=\sum_{k=1}^{n} {k+n \choose n+1}$$ 
 is nothing but coefficient of $x^{n+1}$ in $$(1+x)^{n+1}+(1+x)^{n+2}+(1+x)^{n+3}+....+(1+x)^{n+m}$$ $$=(1+x)^{n+1} [1+(1+x)^1+(1+x)^2+(1+x)^3+...+(1+x)^{m-1}]$$
$$=(1+x)^{n+1}\left(\frac{(1+x)^{m}-1}{1+x-1}\right).$$
So S= coefficient of $x^{n+2}$ in $$ [(1+x)^{n+m+1}- (1+x)^{n+1}] ={n+m+1 \choose n+2}. $$
A: Here is a nice combinatorial approach:
Suppose we have to choose $(n+2)$ boys out of $(m+n+1)$ boys which obviously can be done in ${m+n+1} \choose {n+2}$ ways.
Suppose all these boys are of different ages. Also note that each of these groups has a unique youngest memeber.
Now , we are gonna do the same selection but in a different way.
We first arrange the $(m+n+1)$ boys in ascending order of their ages. Then


*

*we count the no of groups of boys which contain the youngest member, which can be done in ${m+n} \choose {n+1}$ ways(because the youngest boy has already been choosen and we have to choose $(n+1)$ more).

*next, we count the no of groups of boys which contain the second-most youngest member, which can be done in ${m+n-1} \choose {n+1}$ ways( note that these groups can not contain the youngest member)

*...

*...

*lastly we count the no of groups of boys which contain the $m$-th youngest memeber, which can be done in ${n+1} \choose {n+1}$ ways (note that we always have to choose atleast one member whose age is less than or equal to the $m$-th youngest member).
Now we sum up all the number of groups to get the final result.
Thus we have proved-

$$ \sum_{k=1}^{m} {{k+n} \choose {n+1}} = {{ m + n + 1 } \choose {n+2}} $$

