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? 

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.
