<s>[INST] Evaluate an increasing sum of binomial coefficients: $\sum_{k=1}^nk\binom{m+k}{m+1}$ I've been working on a problem and got to a point where I need the closed form of 

$$\sum_{k=1}^nk\binom{m+k}{m+1}.$$

I wasn't making any headway so I figured I would see what Wolfram Alpha could do. It gave me this: 
$$\sum_{k=1}^nk\binom{m+k}{m+1} = \frac{n((m+2)n+1)}{(m+2) (m+3)}\binom{m+n+1}{ m+1}.$$
That's quite the nasty formula. Can anyone provide some insight or justification for that answer? 
 [/INST] For  those who  enjoy integrals  here  is another  approach using  the
Egorychev method as  presented in many posts by  @FelixMarin and also by @MarkusScheuer, where we focus on finding an answer  that differs from the approaches that have already been seen.
Suppose we seek to compute
$$S(n,m) = \sum_{k=0}^n k{m+k\choose m+1}.$$
Introduce
$${m+k\choose m+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2}} (1+z)^{m+k} \; dz$$
as well as the Iverson bracket
$$[[0\le k\le n]]
= \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{w^k}{w^{n+1}} \frac{1}{1-w} \; dw.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2}} (1+z)^{m} 
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} \frac{1}{1-w} 
\sum_{k\ge 0} k w^k (1+z)^k
\; dw \; dz.$$
For this to converge we must have $|w(1+z)|<1.$ We get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2}} (1+z)^{m} 
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+1}} \frac{1}{1-w} 
\frac{w(1+z)}{(1-w(1+z))^2}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2}} (1+z)^{m+1} 
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n}} \frac{1}{1-w} 
\frac{1}{(1-w(1+z))^2}
\; dw \; dz.$$
We evaluate the inner integral using the fact that the residues at the
poles sum to zero. The residue at $w=1$ produces
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2}} (1+z)^{m+1} 
\frac{1}{(-z)^2} \; dz
= -\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+4}} (1+z)^{m+1} \; dz
= 0.$$
For the residue at $w=1/(1+z)$ we re-write the inner integral to get
$$\frac{1}{(1+z)^2} \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n}} \frac{1}{1-w} 
\frac{1}{(w-1/(1+z))^2}
\; dw.$$
We thus require
$$\left.\left(\frac{1}{w^{n}} 
\frac{1}{1-w}\right)'\right|_{w=1/(1+z)}
\\ = \left. \left(\frac{-n}{w^{n+1}} \frac{1}{1-w}
+ \frac{1}{w^n} \frac{1}{(1-w)^2}\right) \right|_{w=1/(1+z)}
\\ = -n (1+z)^{n+1} (1+z)/z + (1+z)^n (1+z)^2/z^2.$$
Substituting this  into the outer  integral and flipping signs  we get
two pieces which are
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2}} (1+z)^{m-1} n(1+z)^{n+2}/z \; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+3}} (1+z)^{n+m+1} \; dz
= n\times {n+m+1\choose m+2}.$$
The second piece is
$$- \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+2}} (1+z)^{m-1}(1+z)^{n+2}/z^2 \; dz
\\ = - \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+4}} (1+z)^{n+m+1} \; dz
= - {n+m+1\choose m+3}.$$
It follows that our answer is
$$\left(n - \frac{n-1}{m+3}\right) {n+m+1\choose m+2}
= \frac{nm+2n+1}{m+3} {n+m+1\choose m+2}.$$
Remark. Being rigorous we also verify that the residue at infinity
in the calculation of the inner integral is zero. We get
$$-\mathrm{Res}_{w=0} \frac{1}{w^2}
w^{n} \frac{1}{1-1/w} \frac{1}{(1-(1+z)/w)^2}
\\ = - \mathrm{Res}_{w=0} 
w^{n-2} \frac{w}{w-1} \frac{w^2}{(w-(1+z))^2}
= - \mathrm{Res}_{w=0} 
\frac{w^{n+1}}{w-1} \frac{1}{(w-(1+z))^2}.$$
There is  certainly no pole  at zero here  and the residue is  zero as
claimed (the  term $1+z$ rotates in  a circle around the  point one on
the real axis  and with $\epsilon \lt 1$ it is  never zero). This last
result could  also be obtained  by comparing degrees of  numerator and
denominator.
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