Find a closed form of $\displaystyle\sum_{n=1}^N n {n \choose k}$.
1) Firstly, is it valid to simplify this equation to: $\sum_{n=k}^N n {n \choose k}$ because ${n \choose k} = 0$ for $n < k$?
2) Is generating functions the right approach to solve this problem? Or should I simply do some algebra?
On
Consider a group of $N+1$ people labeled $1,2,\ldots,N,N+1$. We want to choose $k+1$ of them to form a committee in the following manner, as well as a secretary which may or may not belong to the committee. The committee consists of one president and other members. The president's label must be greater than the labels of all other members and the label of the secretary.
If $n+1$ is the label of the president, then there are $n$ ways to choose the secretary. There are still $k$ members left to be selected among the $n$ people $1,2,\ldots,n$, which can be done in $\displaystyle \binom{n}{k}$ ways. Hence, there are in total $$\sum_{n=0}^N\,n\,\binom{n}{k}$$ ways to decide a committee and its secretary.
On the other hand, consider the two cases. The first case is that the secretary belongs in the committee. Then, there are $k+1$ people to be selected from $N+1$ people, and the person with the largest label is automatically the president. Then, among the remaining $k$ people, one must be chosen to be the secretary. Therefore, there are $$k\,\binom{N+1}{k+1}$$ ways to complete this case. In the second case where the secretary does not belong in the committee, there are $k+2$ people to be selected from $N+1$ people to form the committee and its external secretary, where the one with the largest label is the president. Then, amongst $k+1$ people, one is chosen to be the secretary. Ergo, we can deal with this case in $$(k+1)\,\binom{N+1}{k+2}$$ ways. Hence, there are in total $$k\,\binom{N+1}{k+1}+(k+1)\,\binom{N+1}{k+2}$$ ways to finish the job.
This shows that $$\sum_{n=0}^N\,n\,\binom{n}{k}=k\,\binom{N+1}{k+1}+(k+1)\,\binom{N+1}{k+2}\,.$$ Note that this answer coincides with the answer that can be obtained from Greg Martin's hint, namely, $$k\,\binom{N+1}{k+1}+(k+1)\,\binom{N+1}{k+2}=(k+1)\,\binom{N+2}{k+2}-\binom{N+1}{k+1}\,.$$
On
Use the method of generating functions and snake oil. We wish to evaluate $\sum_{n=0}^Nn\binom{n}{k}$. To this end put $F(z)=\sum_{N=0}^\infty\left(\sum_{n=0}^Nn\binom{n}{k}\right)z^N$. Interchange the order of summation (formally) to get that $$ \begin{align} F(z) &=\sum_{n=0}^\infty n\binom{n}{k}\sum_{N=n}^\infty z^N\\ &=\frac{1}{1-z}\sum_{n=0}^\infty n\binom{n}{k}z^n\\ &=\frac{1}{1-z}(zD)\left(\frac{z^k}{(1-z)^{k+1}}\right)\\ &=\frac{z^k}{(1-z)^{k+3}}(k+z)\tag{0}. \end{align} $$ At this point we wish to extract the coefficient of $z^N$ of $F$. To this end, we use the fact that for integers $m\geq 1$ it is the case that $(1-z)^{-m}=\sum_{n=0}^\infty \binom{m+n-1}{n}z^n$, use $(0)$ and linearity of coefficient extraction to write $$ \begin{align} [z^N](F(z))&=k[z^{N-k}]\left(\frac{1}{(1-z)^{k+3}}\right)+[z^{N-k-1}]\left(\frac{1}{(1-z)^{k+3}}\right)\\ &=k\binom{N+2}{k+2}+\binom{N+1}{k+2}. \end{align} $$ Hence $$ \sum_{n=1}^Nn\binom{n}{k}=k\binom{N+2}{k+2}+\binom{N+1}{k+2}. $$ which agrees with Batominovski's answer after expanding and applying Pascal's identity.
Hint: $$ \sum_{n=k}^N n {n \choose k} = \sum_{n=k}^N (n+1) {n \choose k} - \sum_{n=k}^N {n \choose k} = \sum_{n=k}^N (k+1) {n+1 \choose k+1} - \sum_{n=k}^N {n \choose k} $$ and continue from there....