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


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?

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marked as duplicate by lab bhattacharjee calculus Aug 24 '14 at 16:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The standard proofs are outlined here. – Ahaan S. Rungta Aug 24 '14 at 16:07
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

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

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It is interesting to consider some simple numerical examples here to get a more intuitive feel of the formula. A bar above the number indicates a negative sign.


$$1\quad \bar{2}\quad 1$$


$$1\quad \bar{3}\quad 3\quad \bar{1}$$


$$1\quad \bar{4}\quad 6\quad \bar{4}\quad 1$$


$$1\quad \bar{5}\quad 10\quad \overline{10}\quad 5\quad\bar{1}$$

From the above it is clear that:

  • for odd $n$, a coefficient has the opposite sign of its "mirror image", i.e. $${n\choose k}=-{n\choose n-k}$$ (e.g. 1, -1; -3, 3) thus cancelling out pairwise.

  • for even $n$, this does not occur as a coefficient has the same sign as its "mirror image" (e.g. 1, 1; -4, -4); however, the sum of coefficients in even positions is numerically equal to the sum of coefficients in odd positions, but is negative, thus the sums cancel out (e.g. 1+6+1=4+4, and (1+6+1)+(-4-4)=0)

Thus the formula holds for both odd and even $n$.

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