# Show that $\sum_{m=0}^n(-1)^m\binom nm=0$

Show that $$\sum_{m=0}^n (-1)^m \binom{n}{m} = 0$$

I have first difficulty understanding the summation notation: For example what $$\sum_{m=0}^3 (-1)^m \binom{n}{m}$$ would mean? I suppose that it means: $$\sum_{m=0}^2 (-1)^m \binom{n}{m} = (-1)^0 \binom{2}{0} + (-1)^1 \binom{2}{1} + (-1)^2 \binom{2}{2}$$

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

If it is such case, it shows that the value of $$(-1)^m$$ alternate positive when $$m$$ is even and negative when $$m$$ is odd.

Therefore the sum when $$m$$ is even $$+$$ the sum when $$m$$ is odd $$= 0$$ and we can then factor out the constant $$+1$$ from the first sum and $$-1$$ from the second sum and then QED.

Is this approach correct? How to write this using sum notation when $$m$$ even and when $$m$$ is odd?
Is there a better approach?

Much appreciated.

## 4 Answers

the expansion of $(1+(-1))^n=\sum _0 ^n (-1)^m .(1)^{n-m} {n \choose m}$

• Use ${n\choose m} = {n-1\choose m}+{n-1\choose m-1}$ to see that most of the binomial coefficients cancel. For the rest, use the boundaries ${n\choose 0}= 1={n\choose n}$. – Wuestenfux May 27 '17 at 6:42

This is Newton's Binomial Series:

$$\sum_{m=0}^n\binom{n}{m}(-1)^m = \sum_{m=0}^n\binom{n}{m}(-1)^m1^{n-m} = (-1 + 1)^n = 0^n = 0$$

We have, $$(a+b)^n=\sum\limits_{m=0}^n {n \choose m}a^{n-m}b^m$$

Now, put $a=1, b=-1$

$\sum\limits_{m=0}^n {n \choose m}(-1)^m=(1-1)^n=0$

• Thx for the input, How do you demonstrate $(a+b)^n=\sum\limits_{m=0}^n {n \choose m}a^{n-m}b^m$? – gegu May 27 '17 at 7:08
• That is known as Binomial Expansion. en.wikipedia.org/wiki/Binomial_theorem – Dharmaraj Deka May 27 '17 at 7:11

There is a combinatorial proof of this fact too. Suppose $n\geq 1$ (otherwise it is trivial). Let $N$ be a set of $n$ elements. Let $S_0=\{A\subseteq N\colon |A|\,\text{even}\}$ and $S_1=\{A\subseteq N\colon |A|\,\text{odd}\}$. The identity in question is equivalent to $|S_0|=|S_1|$. Fix an element $a\in N$. Define the map $\varphi\colon S_0\to S_1$ by $$\varphi(A)= \begin{cases} A\cup\{a\}&\text{if a\notin A}\\ A\setminus\{a\}&\text{if a\in A}. \end{cases}$$ It is easy to see that the map $\varphi$ is a bijection (easy to write down an inverse) and hence $|S_0|=|S_1|$.