$\sum_{1\le k\le n,~k\mathrm{~odd}}k\binom{n}{k}=n\cdot2^{n-2}$ for $n>1$

Question

I get some troubles proving this by combinarotics:
$\displaystyle\forall n\in\mathbb{Z}, n>1:\sum_{1\le k\le n,~k\mathrm{~odd}}k\binom{n}{k}=n\cdot2^{n-2}$

Answer 1

HINT: Perhaps the simplest approach is first to use the identity $k\binom{n}k=n\binom{n-1}{k-1}$ to reduce it to

$$\sum_{\substack{1\le k\le n\\k\text{ odd}}}\binom{n-1}{k-1}=2^{n-2}\;.$$

Now notice that the lefthand side can be rewritten

$$\sum_{\substack{0\le k\le n-1\\k\text{ even}}}\binom{n-1}k\;.$$

This last sum counts the even-sized subsets of something.

Answer 2

Suppose you are requested to count the number of ways to form a committee with an odd number of people and one distinguished member out of a group of $n$ people.

One way is to select who are going to be the members and then choose the distinguished one. First select an odd number between 1 and $n$. Afterwards, there are $\binom{n}{k}$ ways to choose a subset of $k$ people, and it is straightforward that the number of ways to choose the special member is $k$. Thus, there are $k\binom{n}{k}$ to choose such a committee with $k$ people. Summing over $k$ gives us the LHS.

Another way to do the same is select the distinguished member and then complete the committee. There are $n$ ways to select the special member. Then you have to choose a subset of the remaining ones with an even number of people (because there is already one person in the committee). It is well-known that the number of subsets of $\{1, \ldots, n\}$ with an even number of elements is $2^{n-1}$. Since you are selecting such subset from a set with cardinal $n-1$, there are $2^{n-2}$ ways. Therefore, there are $n2^{n-2}$ ways to accomplish the task.

Answer 3

Take the derivative of both sides in the identity $$\sum_{0\le k \le n}^n{n\choose k}x^k=(1+x)^{n}$$ to obtain that $$\sum_{1\le k \le n}^nk{n\choose k}x^{k-1}=n(1+x)^{n-1}.$$

Plug in $x=1$ and $x=-1$ and add the two equalities.