proof problem of combination and summation

Question

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

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I don't know why = 0? Please answer my question combination and summation therom is very hard to me.

Answer 1

To prove this yourself (using analysis), start by differentiating

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

with respect to $x$. The result follows from a certain substitution.

Answer 2

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

We need $ S= 1\binom{n}{1} -2 \binom{n}{2} + \cdots + (-1)^{n-1} n \binom{n}{n} $.

In equation (1), differentiate both sides with respect to $x$.

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

Now substitute $x=-1$ in equation (2).

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

$\Rightarrow$ $$S=0=1 \binom{n}{1} -2 \binom{n}{2} + \cdots +n (-1)^{n-1} \binom{n}{n}=\sum_{k=0}^n (-1)^{k-1}k \binom{n}{k}$$

Answer 3

Another alternative is to use identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$

$$ \begin{aligned} \sum_{k=0}^{n}{(-1)^{k-1}\ k\binom{n}{k}}&=n\sum_{k=1}^{n}{(-1)^{k-1}\binom{n-1}{k-1}}\\ &=n(1-1)^{n-1}\\ &=0 \end{aligned} $$