Sum of $\sum\limits_{k=0}^{n}(-1)^{k-1}k\binom{n}{k}$

Question

Can someone give me a hint for solving this? $$\sum_{k=0}^{n}(-1)^{k-1}k\dbinom{n}{k}$$ I tried to manipulate the combinatoric formulas, but it didn't get me anywhere.

Thanks

Answer 1

By the Binomial Theorem we know that

$$(1-x)^n=\sum_{k=0}^n\binom{n}{k}(-1)^k x^k$$

Differentiate the RHS with respect to $x$ leads to

$$\begin{align} \frac{d}{dx}\left(\sum_{k=0}^n\binom{n}{k}(-1)^k x^k\right)=\sum_{k=1}^n\binom{n}{k}(-1)^k k\cdot x^{k-1}&=-\sum_{k=0}^n\binom{n}{k}(-1)^{k-1} k\cdot x^{k-1}\\ &=-(-n(1-x)^{n-1}) \end{align}$$

where the last sum equals the expression $n(1-x)^{n-1}$ which can be obtained by differentiate $(1-x)^n$ directly. For the case $x=1$ we get

$$-\sum_{k=0}^n\binom{n}{k}(-1)^{k-1} k\cdot 1^{k-1}=n(1-1)^{n-1}\Rightarrow\sum_{k=0}^n(-1)^{k-1} k\cdot \binom{n}{k}=-n(1-1)^{n-1}=0$$

Answer 2

Hint:

$$(1-x)^n=\binom{n}{0}-\binom{n}{1}x+\binom{n}{2}x^2-\cdots+(-1)^n\binom{n}{n}x^n$$

Differentiate both side w.r.t. $x$

$$n(1-x)^{n-1}\cdot(-1)=-\binom{n}{1}+2\binom{n}{2}x-\cdots+(-1)^nn\binom{n}{n}x^{n-1}$$

$$n(1-x)^{n-1}=\binom{n}{1}-2\binom{n}{2}x+\cdots-(-1)^nn\binom{n}{n}x^{n-1}$$

now put $x=1$

$$0=\binom{n}{1}-2\binom{n}{2}+\cdots-(-1)^nn\binom{n}{n}$$

$$\sum_{k=0}^{n}(-1)^{k-1}k\dbinom{n}{k}=0$$

Answer 3

Let $n\geq 2$ be an integer. Define $S:=\Big\{(a,A)\,\Big|\,a\in A\subseteq \{1,2,\ldots,n\}\Big\}$. There exists a fixed-point-free involution $f$ on $S$ defined by $$f(a,A)=\begin{cases}\big(a,A\triangle \{1\}\big)\,,&\text{if }a\neq 1\,,\\ \big(a,A\triangle\{2\}\big)\,,&\text{if }a=1\,.\end{cases}$$ Here, $\triangle$ is the symmetric difference operator.

For $j\in\{0,1\}$, let $S_j\subseteq S$ be the subset of $S$ containing $(a,A)$ such that $|A|\equiv j\pmod{2}$. Clearly, $f(S_0)=S_1$. This proves that $|S_0|=\big|f(S_0)\big|=|S_1|$. In other words, $$\sum_{k=0}^n\,(-1)^{k-1}\,k\,\binom{n}{k}=|S_1|-|S_0|=0\,.$$

For $n=1$, the sum is not zero (i.e., $\displaystyle\sum_{k=0}^n\,(-1)^{k-1}\,k\binom{n}{k}=1$ for $n=1$). For $n=0$, the sum is zero. Thus, $$\sum_{k=0}^n\,(-1)^{k-1}\,k\,\binom{n}{k}=\delta_{n,1}\,,$$ where $\delta$ is the Kronecker delta.

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More generally, we have $$\sum_{k=0}^n\,(-1)^k\,\binom{k}{r}\,\binom{n}{k}=(-1)^r\,\delta_{n,r}$$ for all nonnegative integers $n$ and $r$. You can prove this algebraically by noting that $$\binom{k}{r}\,\binom{n}{k}=\binom{n}{r}\,\binom{n-r}{k-r}$$ for integers $n$, $k$, and $r$ such that $n\geq k\geq r\geq 0$. You can also prove this result combinatorially for $n\geq r+1$ by considering $$S:=\Big\{(A,B)\,\Big|\,A\subseteq B\subseteq \{1,2,\ldots,n\}\text{ and }|A|=r\Big\}\,.$$
Then, define a fixed-point-free involution $f$ on $S$ via $$f(A,B):=\Biggl(A,B\triangle \Big\{\min\big(\{1,2,\ldots,r+1\}\setminus A\big)\Big\}\Biggr)$$ for all $(A,B)\in S$. For $j\in\{0,1\}$, let $S_j\subseteq S$ be the subset of $S$ containing $(A,B)$ such that $|B|\equiv j\pmod{2}$. Then, show that $|S_0|=|S_1|$.

Answer 4

Thank you all for your answers

Can we solve this without differentiating like that? :

$\sum_{k=0}^{n}(-1)^{k-1}k\dbinom{n}{k}= n\sum_{k=0}^{n}(-1)^{k-1}\dbinom{n-1}{k-1}$

and $(x-1)^{n-1}$ = $\sum_{k=0}^{n-1}\dbinom{n-1}{k}(-1)^{k}x^{n-1-k}$=$\sum_{k=1}^{n}\dbinom{n-1}{k-1}(-1)^{k-1}x^{n-k}$

And for $x=1$, $$\sum_{k=0}^{n}(-1)^{k-1}k\dbinom{n}{k} = n\sum_{k=1}^{n}\dbinom{n-1}{k-1}(-1)^{k-1} = n(1-1)^{n-1} = 0$$