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Induction Proof - show 2^n = sum of nCi (i = 0 to n)

  1. Jul 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi everyone,
    In my assignment I've been asked to show that 2^n = Ʃ(nCi) i from 0 -> n
    ie: 2^n = nC0 + nC1 + .... + nCn and I have to do this by induction and then also by a combinatorial argument.


    2. Relevant equations
    Right now I'm just working on the induction part.
    BASE CASE: n = 0
    ASSUME: 2^n = ƩnCi
    PROVE: 2^(n+1) = Ʃ(n+1)Ci

    3. The attempt at a solution
    I've shown the base case, 2^0 = 1 and 0C0 = 1
    Then I assumed that 2^k = ƩkCi (i from 0->k)

    Now I've expanded (n+1)Ci = (n+1)C0 + (n+1)C1 +... + (n+1)C(n+1)
    I know there must be a way to do this simply using algebra but I'm totally stumped.

    I tried saying that 2^k+1 = 2(2^k) which we know from the assumption to be 2*(ƩkCi) but that doesn't seem to help me out.

    If anyone could give me a hint which direction to go with the algebra I would really appreciate it!
     
  2. jcsd
  3. Jul 29, 2012 #2

    Ray Vickson

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    Look up Pascal's triangle http://en.wikipedia.org/wiki/Pascal's_triangle (yes, I'm serious!).

    RGV
     
  4. Jul 29, 2012 #3
    I'm still having some trouble with how the algebra works here :(. I know how to show it using the binomial theorem, but our prof explicitly asks us to prove it without using the binomial theorem. Is there a way to do it using the definition of nCk = n!/(n-k)!k! ?? I tried re-writing the sum of (k+1)Ci as (k+1)!/(k+1-i)!i! .. is this a good approach?
     
  5. Jul 29, 2012 #4

    Ray Vickson

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    Did you read my previous response?

    RGV
     
  6. Jul 29, 2012 #5
    I did yes, my prof was also actually talking about Pascal's triangle in class. I'm just having trouble relating the two I think. Their examples seem to focus around the binomial theorem.
     
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