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Math Help - Proof by Induction

  1. #1
    Junior Member
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    Proof by Induction

    Hey guys,

    Ive always have trouble with Inductions no matter how many Ive attempted to do. Please hint/help me please thanks.

    f1^2 + f2^2 + f3^2....fn^2 = fn * fn+1
    Im so lost in this one, I cant even figure out the base case....

    2nd problem
    is a summation problem

    summation of i = 1 to n
    i^2 = n(n+1)(2n+1)/6

    base case of 1 works = 1^2 = 1 and summation = 6/6 = 1

    assume p(n) is true for n = k
    = k(k+1)(2k+1)/6

    n = k+1

    k(k+1)(2k+1)/6 + k+1

    = k(k+1)(2k+1)/6 + 6k+6 = 2k3+2k^3+7k+6...Im stuck here.

    Thanks in advance
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, ff4930!

    You left out a huge chuck of the first problem . . .


    We have the Fibonacci Sequence: 1, 1, 2, 3, 5, 8, 13, ...
    . . where: . F_1 = 1,\;F_2 =1, and: . F_n \:=\:F_{n-1} + F_{n-2}

    Each term is the sum of the preceding two terms.

    Prove: . F_1^2 + F_2^2 + F_3^2 + \cdots + F_n^2 \:=\:F_n\cdot F_{n+1}

    Verify S(1)\!:\;\;F_1^2 \:=\;F_1\cdot F_2
    . . We have: . 1^2 \:=\:1\cdot1 . . . . . true!

    Assume S(k) is true: . F_1^2 + F_2^2 + F_3^2 + \cdots + F_k^2 \;=\;F_k\cdot F_{k+1}

    We want to show the S(k+1) is true:
    . . F_1^2 + F_2^2 + F_3^2 + \cdots + F_{k+1}^2 \:=\:F_{k+1}\cdot F_{k+2}


    Start with S(k):
    . . F_1^2 + F_2^2 + F_3^2 + \cdots + F_k^2 \;=\;F_k\cdot F_{k+1}

    Add F_{k+1}^2 to both sides:
    . . F_1^2 + F_2^2+F_3^2 + \cdots + F_k^2 + F_{k+1}^2 \;=\;F_k\cdot F_{k+1} + F_{k+1}^2

    \text{Factor: }\:F_{k+1}\underbrace{\left(F_k+F_{k+1}\right)}_{\  text{This is }F_{k+2}} \;=\;F_{k+1}\cdot F_{k+2}


    Therefore, we have derived S(k+1):
    . . F_1^2 + F_2^2 + F_3^2 + \cdots + F_{k+1}^2 \;=\;F_{k+1}\cdot F_{k+2}

    The inductive proof is complete.

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  3. #3
    Junior Member
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    Thank you for replying in such short notice.
    Now to stare at it till I get it!
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