How to prove $\sum_{k=1}^n (-1)^{(k+1)} k \binom nk=0$

Question

How to prove this identity?

$$\sum_{k=1}^n (-1)^{(k+1)} k \binom nk=0$$ This is George Casella statistical inference textbook exercise 1.27 (c). I have no idea to prove.

Strong hint: Use the identity $\binom{n}{k} = \binom{n}{n-k}$. — Yanko, Aug 31 '22 at 21:31
@Jonathen In doubt, always use induction for this kind of problem. — Andreas Tsevas, Aug 31 '22 at 21:36
Alternatively use the identity $(X+1)^n=\sum_{k=0}^{n}\binom{n}{k}X^k$. — Ben, Aug 31 '22 at 21:40
Use the binomial theorem to expand $(1 - x)^n$, differentiate both sides of the resulting equation w.r.t. $x$ and then put $x = 1$. — Rob Arthan, Aug 31 '22 at 21:41
@RobArthan Thank for your idea. I almost succeed. But I get $(-1)^{n-k}$, not the desired $(-1)^{k+1}$ — Jonathen, Aug 31 '22 at 22:20
Multiplied by $(-1)^{2k-n+1}$ both sides? It seems a weird step for me. — Jonathen, Aug 31 '22 at 22:29
It's not really weird: it's just about the symmetries in the binomial theorem. If I'd written $(x - 1)^n$, I think you'd have got straight there. — Rob Arthan, Aug 31 '22 at 22:31
Something is not correct in the identity, because it is not true for $n=1$ .You have to say that $n\geq2$ — , Aug 31 '22 at 22:42
Does this answer your question? Binomial coefficient series $\sum\limits_{k=1}^n (-1)^{k+1} k \binom nk=0$ — , Aug 31 '22 at 23:45
@boojum You are right. Same. And that answers my question. I tried to search, but I didn't search the same question. Sad. otherwise I don't need to ask the quesrtion. — Jonathen, Sep 01 '22 at 12:53

score 4 · Accepted Answer · answered Aug 31 '22 at 23:09

\begin{align} \sum_{k=1}^n (-1)^{k+1} k \binom{n}{k} &= n \sum_{k=1}^n (-1)^{k+1} \binom{n-1}{k-1} \\ &= n \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} \\ &= n (1-1)^{n-1} \\ &= n \cdot 0^{n-1} \\ &= \begin{cases} 1 & \text{if $n=1$} \\ 0 & \text{otherwise} \end{cases} \end{align}

score 2 · Answer 2 · answered Sep 01 '22 at 14:24

We assume $n > 1$. Start with the Binomial Theorem: $$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$$ Differentiate: $$n(1+x)^{n-1} = \sum_{k=1}^n k \binom{n}{k} x^{k-1}$$ Set $x=-1$ and observe that $(-1)^{k-1}=(-1)^{k+1}$: $$\begin{align} n(1-1)^{n-1} &= \sum_{k=1}^n k \binom{n}{k} (-1)^{k-1}\\ 0 &= \sum_{k=1}^n k \binom{n}{k} (-1)^{k+1} \end{align}$$

score 0 · Answer 3 · answered Sep 01 '22 at 00:18

First we notice that the identity is not true for $n=1$. So we prove it for $n\geq\,2$. We use induction and the identity:

$\dbinom{n+1}{k}=\dbinom{n}{k}+\dbinom{n}{k-1}$.

Assume it is true for $n$. Now we write the $n+1$ statement as:

$\sum_{k=1}^{n+1}(-1)^{k+1}k\dbinom{n}{k}+\sum_{k=1}^{n+1}(-1)^{k+1}k\dbinom{n}{k-1}$.

Take the first term and notice that for $k=n+1$ the $n+1$ term is zero, by the definition of the factorial coefficient. Therefore we are left with

$\sum_{k=1}^{n}(-1)^{k+1}k\dbinom{n}{k}$ which is zero by the assumption of the induction.

Now we must prove that the second term is also zero. Set $k-1=m$ and get

$\sum_{m=0}^{n}(-1)^{m+2}(m+1)\dbinom{n}{m}$=$\sum_{m=0}^{n}(-1)^{m}(m+1)\dbinom{n}{m}$=$\sum_{m=0}^{n}(-1)^{m}m\dbinom{n}{m}+\sum_{m=0}^{n}(-1)^{m}\dbinom{n}{m}$ but $-\sum_{m=0}^{n}(-1)^{m+1}m\dbinom{n}{m}=0$

by the assumption of the induction (it makes no difference to add $m=0$).

So we are left with $\sum_{m=0}^{n}(-1)^{m}\dbinom{n}{m}$. But this is just (by Newton's binomial theorem) $[1+(-1)]^{n}=0$.

Thus the $n$-statement implies the $n+1$-statement and we are done!

How to prove $\sum_{k=1}^n (-1)^{(k+1)} k \binom nk=0$

3 Answers3