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Math Help - Proof By Induction #2

  1. #1
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    Proof By Induction #2

    I'm using sigma notation and this is what it breaks down to:

    p(k) is k^3= (k^2(k+1)^2)/4
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  2. #2
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    Quote Originally Posted by Edbaseball17 View Post
    I'm using sigma notation and this is what it breaks down to:

    p(k) is k^3= (k^2(k+1)^2)/4
    What would i do to get p(k+1)
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  3. #3
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    Quote Originally Posted by Edbaseball17 View Post
    I'm using sigma notation and this is what it breaks down to:

    p(k) is k^3= (k^2(k+1)^2)/4
    Are you asking how to show that
    \sum_{i = 1}^n i^3 = \frac{1}{4} \cdot n^2(n+ 1)^2?

    The base case is n = 1:
    \sum_{i = 1}^1 1^3 = 1
    and
    \frac{1}{4} \cdot 1^2(1 + 1)^2 = \frac{1}{4} \cdot 4 = 1
    so this checks out.

    Now we assume the theorem is true for some n = k. That is:
    \sum_{i = 1}^k i^3 = \frac{1}{4} \cdot k^2(k + 1)^2

    So we want to show this for n = k + 1.
    \sum_{i = 1}^{k + 1} i^3 = \sum_{i = 1}^k i^3 + (k + 1)^3 = \frac{1}{4} \cdot k^2(k + 1)^2 + (k + 1)^3

    You need to show that this is the same thing as:
    \frac{1}{4} \cdot (k + 1)^2((k + 1) + 1)^2 =  \frac{1}{4} \cdot (k + 1)^2(k + 2)^2

    -Dan
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  4. #4
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    kinda, but the right side is all over 4, not just the 1
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  5. #5
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    Quote Originally Posted by Edbaseball17 View Post
    kinda, but the right side is all over 4, not just the 1
    sorry, just have to learn how to make all the fancy characters, lol.
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