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$$\sum_{k=0}^{k=n}k\ \binom{n}{k}$$ 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.

TShiong
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iimonchii
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  • If the problem were "Find a closed formula for the sum $\sum_{k = 0}^{k=n}\binom nk$", the expected answer would be $2^n$, as the two expressions yield the same value whatever $n$ is, and it is so-called "closed form", which is a semi-rigorous way of saying "simplified". In your problem the answer will be a bit bigger, and presumably not quite that simple, but that's the kind of answer they are after. – Arthur Jan 10 '23 at 12:08
  • The index is not $k=0.$ The index is $k$ and its starting point is $0.$ Therefore the combination $\binom nk$ iterates from $\binom n0$, next to $\binom n1,$ next to $\binom n2,$ and so forth up to and including $\binom nn$. A "closed formula" implies that only certain well-known functions are used and only a fixed number of terms are involved in the formula. Anything that starts with $\sum_{k=0}^{k=n}$, where $n$ is a variable, cannot be a closed form because the number of terms will depend on $n$ and is not a fixed number of terms. – David K Jan 10 '23 at 18:09
  • Note that if you write the sum using $+\cdots+$ notation, you get $0\binom n0 + 1\binom n1 + 2\binom n2 + 3\binom n3 + \cdots + n\binom nn.$ No variable $k$ is required here; the job of $k$ was to help us see the rule for generating each of the terms in that sum, and once we use it for that purpose we don't need it any more. – David K Jan 10 '23 at 18:13

3 Answers3

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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.

Masacroso
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  • where did $kx^{k}$ come from, is there a formal defintion for closed form? and what was the point of redefining the summation with f(x) – iimonchii Jan 10 '23 at 13:18
  • @iimonchii Your problem is a pretty typical combinatorics problem and I assume an introductory textbook would have you string together a few identities to reach an answer. Masacroso here is instead using some techniques for manipulating "Generating Functions" to work toward the answer. – DotCounter Jan 10 '23 at 18:55
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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.

Markus Scheuer
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  • Thank you i understand alot more, could you please explain comment 2 more as I've never seen or at least recognized that identity. does p/q = n? – iimonchii Jan 10 '23 at 17:48
  • @iimonchii: You're welcome. We have $\binom{p}{q}=\frac{p!}{q!(p-q)!}=\frac{p\cdot (p-1)!}{q\cdot (q-1)!(p-q)!}=\frac{p}{q}\binom{p-1}{q-1}$. – Markus Scheuer Jan 10 '23 at 18:23
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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.

DotCounter
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