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?
Using the integral representation of the binomial coefficient $$\dbinom{s}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{s}}{z^{k+1}}dz$$ we have $$ \sum_{k=1}^{n}k\dbinom{m+k}{m+1}=\frac{1}{2\pi i}\sum_{k=1}^{n}k\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{m+k}}{z^{m+2}}dz $$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{m}}{z^{m+2}}\sum_{k=1}^{n}k\left(1+z\right)^{k}dz $$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{m+n+1}}{z^{m+4}}\left(nz-1\right)dz+\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{m+1}}{z^{m+4}}dz$$ $$ =n\dbinom{m+n+1}{m+2}-\dbinom{m+n+1}{m+3} $$ which is equivalent to your claim. To see that we have the same result, note that holds $$\dbinom{n}{k}=\frac{n+1-k}{k}\dbinom{n}{k-1}.$$
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.
You can prove this by induction.
Here is the induction step: $$ \begin{align*} \sum_{k=1}^{n+1} k \binom{m+k}{m+1} &= \frac{n((m+2)n+1)}{(m+2)(m+3)}\binom{m+n+1}{m+1} + (n+1)\binom{m+n+1}{m+1} \\ &=\frac{(m+n+2)(m(n+1)+2n+3)}{(m+2)(m+3)} \binom{m+n+1}{m+1} \\ &=\frac{(n+1)(m(n+1)+2n+3)}{(m+2)(m+3)} \cdot \frac{(m+n+2)!}{(m+1)!(n+1)!} \\ &= \frac{(n+1)((m+2)(n+1)+1)}{(m+2)(m+3)}\cdot \binom{m+n+2}{m+1}. \end{align*} $$
The series can also be seen as the following. \begin{align} \sum_{k=0}^{n} k \, \binom{m+k}{m+1} &= \frac{1}{m+1} \, \sum_{k=0}^{n} k \, \frac{(m+1)_{k}}{k!} \\ &= \frac{1}{m+1} \, \left[ \sum_{k=0}^{n-2} \frac{(m+1)_{k+2}}{k!} + \sum_{k=0}^{n-1} \frac{(m+1)_{k+1}}{k!} \right] \\ &= (m+2) \, \frac{(n-1) \, \Gamma(m+n+2)}{\Gamma(n) \, \Gamma(m+4)} + \frac{ \Gamma(m+n+2)}{\Gamma(n) \, \Gamma(m+3)} \\ &= \frac{(m+n+1)!}{(n-1)! \, (m+3)!} \, (m \, n + 2n + 1) \\ &= \frac{m \, n + 2n + 1}{m+3} \, \binom{m+n+1}{n-1}. \end{align} Rearranging terms yields the form presented in the proposed problem. The notation used in Pochhammer's notation, namely, \begin{align} (x)_{k} = \frac{\Gamma(x+k)}{\Gamma(x)}. \end{align}
Here is a slightly different variation of the theme. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. This way we can write e.g. \begin{align*} [x^k](1+x)^n=\binom{n}{k}\tag{1} \end{align*}
We obtain \begin{align*} \sum_{k=1}^{n}&k\binom{m+k}{m+1}=\sum_{k=1}^nk[x^{m+1}](1+x)^{m+k}\tag{2}\\ &=[x^{m+1}](1+x)^{m+1}\sum_{k=1}^nk(1+x)^{k-1}\tag{3}\\ &=[x^{m+1}](1+x)^{m+1}D_x\left(\sum_{k=1}^n(1+x)^{k}\right)\tag{4}\\ &=[x^{m+1}](1+x)^{m+1}D_x\left(\frac{1-(1+x)^{n+1}}{1-(1+x)}-1\right)\tag{5}\\ &=[x^{m+1}](1+x)^{m+1}\left(\frac{(nx-1)(1+x)^n}{x^2}+\frac{1}{x^2}\right)\tag{6}\\ &=[x^{m+3}](1+x)^{m+n+1}(nx-1)\tag{7}\\ &=n[x^{m+2}](1+x)^{m+n+1}-[x^{m+3}](1+x)^{m+n+1}\tag{8}\\ &=n\binom{m+n+1}{m+2}-\binom{m+n+1}{m+3}\tag{9} \end{align*}
Comment:
In (2) we apply the coefficient of operator according to (1)
In (3) we use the linearity of the coefficient of operator and split the binomial conveniently
In (4) we introduce the differential operator $D_x:=\frac{d}{dx}$
In (5) we use the formula for the finite geometric series
In (6) we apply the differential operator $D_x$
In (7) we do some simplifications and use the rule \begin{align*} [x^m]x^{-k}A(x)=[x^{m+k}]A(x) \end{align*}
In (8) we use the linearity of the coefficient of operator and apply the rule above again
In (9) we write the expression using binomial coefficients and obtain a result in accordance with the answer of @MarcoCantarini
