<s>[INST] Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let n=0. Then $\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$. Thus for $n=0$ our equation is satisfied.
Inductive Step. Let $n\ge0$. Assume $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$. Now observe that
$$\begin{align*}\sum_{k=0}^{n+1}k{m+k \choose m}&=\\
\sum_{k=0}^{n}k{m+k \choose m}+(n+1){m+n+1\choose m}&=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}+(n+1){m+n+1\choose m} \end{align*}$$
This is where I'm getting stuck... using Pascal's Identity to combine some of these terms seems ideal. However, the factors of $n$ and $n+1$ are making a clever manipulation difficult for me.
 [/INST] Here is a somewhat different approach. We use the coefficient of operator $[x^k]$ to denote  the  coefficient  of $x^k$ in  a  series. This way we can  write  e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}

We obtain  for   $m,n\geq 0$
  \begin{align*}
\sum_{k=0}^n&k\binom{m+k}{m}\\
&=\sum_{k=1}^nk[x^m](1+x)^{m+k}\tag{1}\\
&=[x^m](1+x)^{m+1}\sum_{k=1}^nk(1+x)^{k-1}\tag{2}\\
&=[x^m](1+x)^{m+1}D_x\left(\sum_{k=1}^n(1+x)^k\right)\tag{3}\\
&=[x^m](1+x)^{m+1}D_x\left(\frac{1-(1+x)^{n+1}}{1-(1+x)}\right)\tag{4}\\
&=[x^m](1+x)^{m+1}\left[\frac{n(1+x)^n}{x}-\frac{(1+x)^n}{x^2}+\frac{1}{x^2}\right]\tag{5}\\
&=n[x^{m+1}](1+x)^{m+n+1}-[x^{m+2}](1+x)^{m+n+1}+[x^{m+2}](1+x)^{m+1}\tag{6}\\
&=n\binom{m+n+1}{m+1}-\binom{m+n+1}{m+2}
\end{align*}
  and the  claim  follows

Comment:


*

*In (1) we apply the coefficient of operator.

*In (2) we do a little rearrangement  by using the linearity  of the coefficient  of  operator.

*In (3) we use  the differential operator $D_x:=\frac{d}{dx}$.

*In (4) we use the formula for the finite geometric series.

*In (5) we differentiate the expression.

*In (6) collect terms and apply the formula
\begin{align*}
[x^{p+q}]A(x)=[x^p]x^{-q}A(x)
\end{align*}

*In (7) we select the coefficients of $x^{m+1}$ and $x^{m+2}$. Note the coefficient of $x^{m+2}$ in the last term is zero, since the polynomial $(1+x)^{m+1}$  has degree less than $m+2$.
</s>