What does it mean when the stopping point is equal to the index in a summation? $$\sum_{k=0}^{k=n}k\ \binom{n}{k}$$
Find a closed formula for the sum. The index is $k=0$ and the stopping point is $k=0$ so I am assuming the process is that the combination iterates from $^nC_k$ to $^nC_n$. How would i show that in a closed formula? I am confused on what a closed formula is in the first place. What I am assuming is that it is a formula that represents the summation and gives the same values when substituting in the values. However i am not sure how to go about it.
 A: By definition of $\sum$ you have that
$$
\sum_{k=0}^n f(k)=f(0)+f(1)+\ldots +f(n)
$$
It doesn't matter if $n$ is in the definition of $f$, as $n$ is a constant. Now, for the solution, observe that
$$
\sum_{k=0}^n k \binom{n}{k}=\left[\sum_{k=0}^n \binom{n}{k} kx^k\right]_{x=1}
=\left[\sum_{k=0}^n \binom{n}{k} x\frac{d}{d x}x^k\right]_{x=1}=\left[x\frac{d}{d x}\sum_{k=0}^n \binom{n}{k} x^k\right]_{x=1}=\ldots 
$$
I hope you can follow from here.
A: A closed formula here means a representation without a sum symbol (and without using a bound variable as index variable $k$).

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^nk\binom{n}{k}}&=\sum_{k=1}^nk\binom{n}{k}\tag{1}\\
&=n\sum_{k=1}^n\binom{n-1}{k-1}\tag{2}\\
&=n\sum_{k=0}^{n-1}\binom{n-1}{k}\tag{3}\\
&\,\,\color{blue}{=n2^{n-1}}\tag{4}
\end{align*}
where $n2^{n-1}$ is a closed form representation.

Comment:

*

*In (1) we start with the index $k=1$ since the summand with $k=0$ does not contribute.


*In (2) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


*In (3) we shift the index to start with $k=0$.


*In (4) we apply the binomial theorem.
Hint: Some more information about a closed formula is given in this MSE answer.
A: 
What does it mean when the stopping point is equal to the index in a summation?... The index is k=0 and the stopping point is k=n...

I think you are a bit confused about the way that the sigma notation works.
$$
\sum_{k=0}^{k=n} k \binom{n}{k}
$$
In this example, $k$ is the index and it starts at $0$ and continues up until it reaches $n$, and each $k\binom{n}{k}$ term is evaluated for each allowed value of $k$.  In other words:
$$
\sum_{k=0}^{k=n} k \binom{n}{k}=\Bigg[k \binom{n}{k}\Bigg]_{k=0}+\Bigg[k \binom{n}{k}\Bigg]_{k=1}+\Bigg[k \binom{n}{k}\Bigg]_{k=2} + \ ... + \Bigg[k \binom{n}{k}\Bigg]_{k=n}
$$
$$
\sum_{k=0}^{k=n} k \binom{n}{k}=\Bigg[0 \binom{n}{0}\Bigg]+\Bigg[1 \binom{n}{1}\Bigg]+\Bigg[2 \binom{n}{2}\Bigg] + \ ... + \Bigg[n \binom{n}{n}\Bigg]
$$
The part about finding a "closed formula" is just asking for you to simplify the expression into a form that no longer requires a summation.  This will require the use of multiple different identities and is covered by other users answers very well.
