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I am struggling with computing the following sums:

$$\sum_{k=1}^{n}k\binom{n}{k}=\binom{n}{1}+2\binom{n}{2}+...+n\binom{n}{n}$$

and

$$\sum_{k=0}^{n}\frac{1}{k+1}\binom{n}{k}=\binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+\dots+\frac{1}{n+1}\binom{n}{n}$$

At first, I tried just rewriting the general terms in a form where the Binomial Theorem could be applied, but could not do so. ($k$ should be in the exponent of something with $n-k$ in the exponent of something else.)

Then, for the first sum, I re-wrote it in the following manner:

\begin{align} S &=\binom{n}{n}+\dots+\binom{n}{3}+\binom{n}{2}+\binom{n}{1}\\ &+\binom{n}{n}+\dots+\binom{n}{3}+\binom{n}{2}\\ &+\binom{n}{n}+\dots+\binom{n}{3}\\ &\quad\vdots\\ &+\binom{n}{n} \end{align}

Noting that:

$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\dots+\binom{n}{n}=\sum_{k=0}^{n}\binom{n}{k}1^k1^{n-k}=(1+1)^n=2^n,$$

I can rewrite the first line in $S$ as $2^n-1$ (the $-1$ is there because I'm over-counting the $\binom{n}{0}=1$).

Then I tried to say that if all the "blanks" were filled (if all the lines were like the first line), since there are $n$ lines, then I have $n(2^n-1)$. From this point, we should have:

$$S+\text{blanks}=n(2^n-1)\Longrightarrow S=n(2^n-1)-\text{blanks}$$

However, the problem of finding the value of "blanks" seems as hard as the original problem and I can't get to the right answer of $n2^{n-1}$ (according to WA). In one of my attempts, I ended up with a geometric series but that still didn't work.

I have not spent much time on the second sum as I feel I should be able to do the first one before even tackling the second one. Am I going in the right direction with this? Is there a more intelligent way of working this out?

Thanks in advance for any help/hints.

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marked as duplicate by N. F. Taussig combinatorics Aug 11 '18 at 8:55

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    $\begingroup$ You may just differentiate/integrate $\sum_{k=1}^{n}\binom{n}{k}x^k = (1+x)^n-1$. $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 3:13
  • $\begingroup$ Oh, I really should have thought of that. I tunnel-visioned myself into thinking this should be do-able with just basic counting tricks, not calculus. Thanks for the tip, I will try. $\endgroup$ – orion2112 Aug 11 '18 at 3:17
  • $\begingroup$ It is doable with basic tricks, but the solution via $\frac{d}{dx},\int(\ldots)dx$ is a one-liner. $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 3:26
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One of my preferred methods of proof for these types of identities is to use a combinatorial proof which generally takes the form of describing a counting problem and finding the answer using two different methods thereby proving that the expressions achieved by each answer must be equal.

Suppose we have a committee of $n$ distinct people. We ask, in how many ways may we choose a subcommittee of any size where the subcommittee has one member designated as the leader (including the case of the leader being the only person on the subcommittee by himself)?

  • Counting Method 1

Let us first break into cases based on the number of members on the committee. Since the committee must have a leader, the committee size must be at least $1$ and can be at most $n$. Let $k$ be the number of people on the subcommittee

Next, let us select all of the members of the subcommittee (including the leader). This can be done in $\binom{n}{k}$ ways.

Then, from those people selected, let us choose one of them to be the leader. This can be done in $k$ ways.

Applying multiplication principle and summing over all possible cases gives us the total number of subcommittees possible as being:

$$\sum\limits_{k=1}^nk\binom{n}{k}$$

  • Counting Method 2

Let us before anything else select a person to serve as the leader for the subcommittee. This can be done in $n$ ways.

From the remaining $n-1$ people, let us choose some subset (possibly empty) to act as the followers in the subcommittee. This can be done in $2^{n-1}$ ways.

Applying multiplication principle, we arrive at a total number of subcommittees as being:

$$n2^{n-1}$$


We have as a result the identity:

$$\sum\limits_{k=1}^nk\binom{n}{k}=n2^{n-1}$$

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