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I just got to this question and I became a question mark. I wonder if anyone can help me with this one, because I don't even know how to begin to tackle this problem.

The question:

Prove the formula $\sum_{k=1}^n k\binom{n}{k} = n \cdot 2^{n-1}$ for all integers $n > 0$. We accept a proof based on manipulations of formulas and do not demand a proof by mathematical induction here.

Hint: Write out the first few terms in the sum $\sum_{k=1}^n k\binom{n}{k}$ and factor out $n$. Then use the formula $\sum_{j=0}^m \binom{m}{j} = 2^m$ that is valid for all non-negative integers $m$, particularly $m = n−1$.

Please help. I don't understand it.

NoChance
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haha150
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    Have you tried reading through and understanding the hint? Try rewriting $\binom{n}{k}$ as $\frac{n!}{(n-k)!k!}$. Notice that $k\binom{n}{k}$ you can cancel one of the $k$'s on the bottom. Factor out an $n$... This leaves you with $k\binom{n}{k} = n\cdot \frac{(n-1)!}{(n-k)!(k-1)!}$. Does the term on the right look familiar? In particular if you rewrite $(n-k)$ as $((n-1)-(k-1))$? (all of this was already stated in the hint, just restating it in different/longer words). – JMoravitz Feb 10 '16 at 21:55
  • A combinatorial proof is here: http://math.stackexchange.com/questions/1638943/give-a-combinatorial-proof-to-show-sum-i-1nicn-i-n2n-1 – NoChance Feb 10 '16 at 22:30
  • The question in favor of which I closed it has a very full explanation of a combinatorial proof. The hint is suggesting a different proof, one that makes use of the identity explained in this answer. – Brian M. Scott Feb 11 '16 at 03:32
  • why would you close this when there is no question like this out there, if you show me this question using the same hint then you can keep it closed, ty – haha150 Feb 11 '16 at 21:38

4 Answers4

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Need to follow hint your professor said, calculate a few terms, and factor out n:

$$k\binom nk = k \frac{n!}{k!(n-k)!} = n\frac{(n-1)!}{(k-1)!(n-k)!} = n\frac{(n-1)!}{(k-1)!((n-1) - (k - 1))!} = n\binom {n-1}{k-1}$$

Follow the rest hints to finish the proof.

Now we use the hint $$\sum_{j=0}^m \binom{m}{j} = 2^m$$

$$\sum_{k=1}^n k \frac{n!}{k!(n-k)!} = \sum_{k=1}^n n\binom {n-1}{k-1}$$ $$=n\sum_{k=1}^n\binom {n-1}{k-1}$$ let $ j = k -1$, so $ j = 1 ... (n-1)$, using hint with $m = n -1$ $$=n\sum_{j=0}^{n-1}\binom {n-1}{j} = n 2^{n-1}$$

runaround
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I will give an inductive proof using the fact that $ \sum_{k=0}^n \tbinom n k = 2^n$ as well as Pascal's identity.

The base case is straightforward.

Suppose it holds for some $n$. Then $$\sum_{k=1}^{n+1} k\binom {n+1} k = n+1 + \sum_{k=1}^{n} k\binom {n} {k-1} + \sum_{k=1}^{n} k\binom {n}{k} = n + 1 + \sum_{k=1}^{n} (k-1)\binom {n} {k-1} + \sum_{k=1}^{n} \binom {n} {k-1} + \sum_{k=1}^{n} k\binom {n}{k} = (n+1) + (n \cdot 2^{n-1} - n) + (2^{n} - 1) + (n \cdot 2^{n-1}) = n\cdot2^n + 2^n = (n+1)2^{(n+1)-1}$$.

hence it holds for $n+1$.

This completes the proof.

MathematicsStudent1122
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Hint:

Can you calculate $$\sum_{k=1}^nk\binom nk x^{k-1}?$$

Bernard
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I a giving a combinatorial proof.

Consider a set $S$ with $n$ elements. Now, calculate the number of elements in every subsets of this set $S$. So, in the subsets containing $k$-elements, there are $\color{blue}{k\binom nk}$ elements in total. So, for $0\le k\le n$, the number of all the elements in all the possible subsets is $$\color{red}{{\sum\limits_{k=1}^nk\binom nk}}\tag 1.$$ Now, count in a different way. Count the total number of elements of a subset along with its complementary subset.

Say $A$ be a subset of $S$, then #$A+$#$A^c=\color{blue}{n}$. And earlier, there was $2^n$ subsets. So, while counting the number of subsets along with their complements, we should have $2^n/2=\color{blue}{2^{n-1}}$ subsets to count, each containing $n$ elements.

So, in this way, you get, the number of all the elements in all the possible subsets is $$\color{red}{n\times 2^{n-1}}.\tag 2$$

In both cases, we counted same quantity, so, the results should be equal. So, from $(1)$ and $(2)$ we get, $$\sum k\binom nk=n\times 2^{n-1}.$$

  • Oh, a combinatorial proof link was posted in comments. But when I started writing this answer, that was not still posted then. –  Feb 10 '16 at 22:41
  • Maybe a stupid question but is this the full proof and would it be sufficient? – haha150 Feb 10 '16 at 23:53
  • @haha150, yes, combinatorial proofs are well acclaimed and no doubt, extremely beautiful. And this is more than sufficient. But if it your exam question, then I would suggest not to do it there, unless you are much confident about this. And also examiners may not have much time to see this. –  Feb 12 '16 at 05:33