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Formula for the n-th row of Pascal's Triangle

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a formula for the sum of the elements of the nth row of Pascal's Triangle


    2. Relevant equations
    C(n,r) = [C(n-1,r-1) + C(n-1,r)]
    C(n,0) = C(n,n) = 1


    3. The attempt at a solution

    I started with the summation of the elements in the rows n
    [itex]\sum^{n}_{r=0} C(n,r)[/itex]
    = [itex]C(n,0) + \sum^{n-1}_{r=1} C(n,r) + C(n,n)[/itex]
    = [itex]C(n,0) + \sum^{n-1}_{r=1} [C(n-1,r-1) + C(n-1,r)] + C(n,n)[/itex]
    = [itex]C(n,0) + \sum^{n-1}_{r=1} C(n-1,r-1) + \sum^{n-1}_{r=1} C(n-1,r) + C(n,n)[/itex]
    = [itex]C(n,0) + \sum^{n-2}_{r=0} C(n-1,r) + \sum^{n-1}_{r=1} C(n-1,r) + C(n,n)[/itex]
    = [itex]C(n,0) + [C(n-1,0) + \sum^{n-2}_{r=1} C(n-1,r)] + [\sum^{n-2}_{r=1} C(n-1,r) + C(n-1,n-1)] + C(n,n)[/itex]

    = [itex]2 \cdot C(n-1,0) + \sum^{n-2}_{r=1} C(n-1,r) + \sum^{n-2}_{r=1} C(n-1,r) + 2 \cdot C(n-1,n-1)[/itex] since C(n,0) = 1 = C(n-1,0) and C(n,n) = 1 = C(n-1,n-1)
    =[itex]2 \cdot \sum^{n-1}_{r=0} C(n-1,r)[/itex]

    This gives the intuitively correct response that the sum of the nth row is twice that of the (n-1)th row. How can I go from here to sum = 2^n?

    Thank you!
     
  2. jcsd
  3. Jan 22, 2013 #2

    jbunniii

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    Do you know the binomial theorem? What is a formula for [itex](x+y)^n[/itex] in terms of [itex]C(n,r)[/itex]?
     
  4. Jan 22, 2013 #3

    jbunniii

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    This should be a simple induction. What is the sum of the 1st row?
     
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