Alternating sum of binomial coefficients is equal to zero

Question

Prove without using induction that the following formula:$$\sum_{k=0}^n (-1)^k\binom{n}{k}=0$$ is valid for every $n\ge1$.

Progress

For each odd $n$ we can use the identity:$$\binom{n}{k}=\binom{n}{n-k}$$ In fact all terms equidistant from the end points are opposite. My question is: if $n$ is even how can we prove it?

Related : http://math.stackexchange.com/questions/634956/2n-c-0c-1-dotsc-n/634959#634959 — lab bhattacharjee, Aug 24 '14 at 16:09
Depending on your perspective, this fact shows that the Euler characteristic of the $n$-torus $T^n = (S^1)^n$ is zero, or it can be proven using that $\chi(T^n) = 0$. — Phillip Andreae, Aug 24 '14 at 16:18

score 42 · Answer 1 · answered Aug 24 '14 at 14:25

42

Set $a=1,b=-1$ in the Binomial Expansion formula for non-negative intger $n$ $$(a+b)^n=\sum_{k=0}^n\binom nk a^{n-k}b^k$$

answered Aug 24 '14 at 14:25

lab bhattacharjee

279,016

1

So simple yet so efficient. – Sirswagger21 Jun 29 '21 at 20:44

Alternating sum of binomial coefficients is equal to zero

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