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Proof By Induction: Fibonacci Sequence

  1. Feb 18, 2012 #1

    dba

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    1. The problem statement, all variables and given/known data
    [itex]\sum_{i=0}^{k} {k \choose i}f_{n+i} = f_{n+2k} [/itex]
    2. Relevant equations
    Definition:
    [itex] f(0)=0 ; f(1)=1 [/itex]
    [itex]f(n)=f_{n-1} + f_{n-2}[/itex] for [itex] n>=2[/itex]


    3. The attempt at a solution
    I searched a basis for which the statement is true:
    n=2 and k=1
    [itex]\sum_{i=0}^{1} {1 \choose i}f_{2+i} = f_{2+2(1)} [/itex]
    i=0: [itex]{1 \choose 0}f_{2+0} + {1 \choose 1}f_{3} = 3[/itex]

    and

    [itex]f_{n+2k} = f_{2+2(1)} = f_{4} = 3[/itex]

    I assume for k-1
    [itex]\sum_{i=0}^{k-1} {k-1 \choose i}f_{n+i} = f_{n+2(k-1)} [/itex]

    I need to show
    [itex]\sum_{i=0}^{k} {k \choose i}f_{n+i} = f_{n+2k} [/itex]

    Normally I would take the Left-hand-side(LHS) and
    [itex]\sum_{i=0}^{k} {k \choose i}f_{n+i} = \sum_{i=0}^{k-1} {k-1 \choose i}f_{n+i} +k [/itex]

    I do not know how to write the summation notation in a way that I could get to the Right-hand-side(RHS) which would be f_{n+2k}

    Thanks for any help.
     
  2. jcsd
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