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I want to know if I can get some help with this proof. I tried, but I just cannot seem to get $2^{k}$. It states that,

For $k \in \mathbb{Z}_{\ge 0}$, $$\sum^{k}_{m=0}\binom{k}{m} = 2^k$$

Thank You.

  • 1
    Are you familiar with the binomial formula? Because once you know (or have proven) the binomial formula, what you want to prove should follow pretty quickly.2011-03-17
  • 2
    How do you **define** the binomial coefficients? There are many (equivalent) ways and, depending on the way you choose, the proof you want is not the same.2011-03-17
  • 0
    @MarcvanLeeuwen Based on [this meta discussion](http://meta.math.stackexchange.com/questions/10615/when-using-summation-tag-should-we-remove-sequences-and-series) I think that this should be tagged ([tag:summation]) and the tag ([tag:sequences-and-series]) should not be used. (As far as I know, the later is intended for infinite sums. Although many questions about finite sums are tagged with this tag, since they were asked before ([tag:summation]) tag was created.)2014-01-09
  • 0
    About closing as a duplicate: One of the questions specifically asks for a proof about induction, the other does not have this restriction.2014-01-09

4 Answers 4