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Proof by mathematical induction

  1. Sep 6, 2014 #1
    Hello,

    I need to prove the following:
    [tex]\sum_{i=0}^n\binom{n}{i} = 2^n[/tex]
    by using something called mathematical induction. I understand, somewhat, what it is - we propose a statement and show that is true for n=1, then we assume that the statement is true for all [itex]n \in \mathbb{N}[/itex], which should also mean that the statement is true for n+1. This is what I have written down:

    [tex]\binom{n+1}{0} + \binom{n+1}{1}+ . . .+ \binom{n+1}{n-1}+ \binom{n+1}{n}+ \binom{n+1}{n+1} = 2^{n+1} [/tex]
    which is
    [tex]1 + (n+1) + \frac{n(n+1)}{2!}+ . . .+ \frac{n(n+1)}{2!}+ (n+1)+ 1 = 2^{n+1}[/tex]
    I can see a symmetry and I thought about calculating half and showing that it equals 2n, but that idea quickly died, since I don't know where the "half point" is in the sum.

    What should I do?
    Thanks in advance!
     
    Last edited: Sep 6, 2014
  2. jcsd
  3. Sep 6, 2014 #2

    Ray Vickson

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    Given the truth of the statement for ##n=k##, you need to prove it for ##n= k+1##. To do that, you need to know how ##\binom{n+1}{r}## relates to ##\binom{n}{s}##. Use Pascal's Triangle; look it up if you have not heard of it.
     
  4. Sep 6, 2014 #3
    I looked it up Here and saw the Pascal's rule, according to which I could state that:
    [tex]\binom{n+1}{k} = \binom{n}{k-1}+ \binom{n}{k}, n \geq 0 \land k \in [1, n][/tex]
    (k can't be zero, right? If it is then nCk-1)

    If I plug in k=n I get:
    Left side:
    [tex]\frac{(n+1)!}{n!}= n+1[/tex]
    Right side:
    [tex]\frac{n!}{(n-1)!}+ \frac{n!}{n!}= n+1 [/tex]
    Cannot understand how this helps me come to the conclusion in the first statement(the original assignment). What exactly does that equality show?
     
  5. Sep 6, 2014 #4

    Fredrik

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    It doesn't look like you're trying to use induction. The idea between induction is this: Suppose that for each natural number n, P(n) is a statement about n. You can prove that P(n) is true for all n, by proving only two statements:

    1. P(0)
    2. For all n, if P(n) then P(n+1).

    In your case, P(n) is the statement ##\sum_{i=1}^n \binom n i=2^n##. You need to prove P(0), i.e. that ##\sum_{i=0}^0 \binom 0 i =2^0##. This is of course easy. Then you let n be an arbitrary natural number and try to prove that if P(n) then P(n+1). So you have to use ##\sum_{i=0}^n \binom n i =2^n## to prove ##\sum_{i=0}^{n+1}\binom{n+1}{i}=2^{n+1}##.
     
    Last edited: Sep 6, 2014
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