I have studying my maths book induction chapter and I found things to solve this but I am failed, somebody help me to solve this problem by simple method of mathematical induction. $$2^n = {n \choose 0} +{n \choose 1} + \cdots {n \choose n}$$ How to prove it by mathematical induction.
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2Have you proven it for the base case? – Ken May 29 '15 at 21:36
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use that $2^{n}=(1+1)^{n}$ – Dr. Sonnhard Graubner May 29 '15 at 21:43
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There are better ways that do not use induction directly. But one direct way is to use $\binom{k+1}{r}=\binom{k}{r}+\binom{k}{r-1}$ (Pascal Identity). – André Nicolas May 29 '15 at 21:46
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$${n \choose 0} +{n \choose 1} + \cdots + {n \choose n}$$ is the number of ways to choose any number of things out of $n$ things, which clearly explains why this must be equal to $2^n$ (since any of the things may or may not be chosen). It depends exactly on the provided definitions how you may or may not prove this.
For the definition you could use Newton's binomium, then you could kind of use induction going down the triangle. It is also possible to use the defintion
$${n \choose k}=\frac{n!}{k!(n-k)!}$$
but this is harder.
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