Math 2.0

Combinatorics Theorem

Alternating sum of binomial coefficients

Builds on Binomial coefficient

Statement

Let nn be a positive integer. Then ∑k=0n(−1)k(nk)=0.\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0.

Remarks

For n=0n = 0 the sum is 11, so the hypothesis n≥1n \ge 1 is necessary. For n=4n = 4: 1−4+6−4+1=01 - 4 + 6 - 4 + 1 = 0. The identity is the binomial theorem with x=−1x = -1 and y=1y = 1. It says that a nonempty finite set has as many subsets of even size as of odd size.

Reading path

To understand this page from the ground up, read these 2 pages in order. The foundations come first.

Show the path as a graph
Binomial coefficient, definition, no evidence levelBinomial coefficientdefinitionAlternating sum of binomial coefficients, statement, no evidence levelAlternating sum of bi…statement
Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  1. Binomial coefficient definition
  2. Alternating sum of binomial coefficients statement

Further reading

  1. Web https://en.wikipedia.org/wiki/Binomial_coefficient, Sums of the binomial coefficients
About this page
Tag0PQQGFAWSH
URIm2://0PQQGFAWSH
Kindstatement, theorem
WritingW0 stub
MSC05A10
StatusThis copy of the site has no build data. Run m2 status 0PQQGFAWSH.
UsesBinomial coefficient

Markdown · JSON · Edit on GitHub

What this page needs for W1

  • an inventory row: the row of the inventory that the node accounts for (0 of 1)

The writing levels say how complete the text of a page is. They do not measure its truth.

What links here

No page links to this page yet.