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Thread: Mathematical Induction

  1. #1
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    Mathematical Induction

    Hello!
    I've got this problem.


    18. Consider the following four equations:

    1) 1 = 1
    2) 2+3+4 = 1+8
    3) 5+6+7+8+9 = 8+27
    4) 10+11+12+13+14+15+16 = 27+64
    Conjecture the general formula suggested by these four equations, and prove your conjecture.

    Any help would be appreciated
    Thanks!
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  2. #2
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    here's my working
    Attached Thumbnails Attached Thumbnails Mathematical Induction-out0071.jpeg  
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by BACONATOR View Post
    Hello!
    I've got this problem.


    18. Consider the following four equations:

    1) 1 = 1
    2) 2+3+4 = 1+8
    3) 5+6+7+8+9 = 8+27
    4) 10+11+12+13+14+15+16 = 27+64
    Conjecture the general formula suggested by these four equations, and prove your conjecture.

    Any help would be appreciated
    Thanks!
    Conjecture: $\displaystyle \forall n \in \mathbb{N}$

    $\displaystyle \sum_{r=(n-1)^2+1}^{n^2} r = (n-1)^3+n^3$

    and you should not need induction to prove this as the left hand side is an
    arithmetic progression.

    RonL
    Last edited by CaptainBlack; Feb 22nd 2008 at 02:11 AM.
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  4. #4
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    Hello, BACONATOR!

    Find the general statement is the biggest task . . .


    18. Consider the following four equations:

    $\displaystyle \begin{array}{cccccc}
    1) & 1 & = & 1 & = & 0^3+1^3\\
    2) & 2+3+4 &= & 1+8 & = & 1^3+2^3\\
    3) & 5+6+7+8+9 & = & 8+27 &=& 2^3+3^3\\
    4) & 10+11+12+13+14+15+16 & = & 27+64 & = & 3^3+4^3
    \end{array}$

    Conjecture the general formula suggested by these four equations,
    and prove your conjecture.
    Here's the pattern that I saw . . .

    Equation 2: sum of 3 consecutive integers up to 2, equals 1 + 2

    Equation 3: sum of 5 consecutive integers up to 3, equals 2 + 3

    Equation 4: sum of 7 consecutive integers up to 4, equals 3 + 4


    My conjecture:

    Equation $\displaystyle n$: sum of $\displaystyle 2n-1$ consecutive integers up to $\displaystyle n^2$, equals $\displaystyle (n-1)^3 + n^3$


    After a bit of algebra, I found that equation $\displaystyle n$
    . . has a sum that begins with $\displaystyle n^2-2n+2$

    The sum is an arithmetic series with first term, $\displaystyle a \:= \:n^2-2n+2$,
    . . common difference, $\displaystyle d = 1$, and $\displaystyle 2n-1$ terms.

    This sum is: .$\displaystyle S \;=\;\frac{2n-1}{2}\left[2(n^2-2n+2) + 2n(1)\right]$

    . . which simplifies to: .$\displaystyle 2n^3 - 3n^2 + 3n - 1 \;=\;(n-1)^3 + n^3$

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  5. #5
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    Re: Mathematical Induction

    Hello, BACONATOR!

    $\displaystyle \text{18. Consider the following four equations:}$

    $\displaystyle \begin{array}{cccccc}(1) & 1 &=& 0^3 + 1^3 \\(2) & 2+3+4 &=& 1^3+2^3 \\ (3) & 5+6+7+8+9 &=& 2^3+3^3 \\ (4) & 10+11+12+13+14+15+16 &=& 3^3+4^3 \end{array}$

    $\displaystyle \text{Conjecture the general formula suggested by these four equations,}$
    . . $\displaystyle \text{and prove your conjecture.}$

    The left side is an arithmetic series.
    . . Its first term is: $\displaystyle a = k^2 - 2k+2$
    . . The common difference is: $\displaystyle d = 1$,
    . . The number of terms is: $\displaystyle n \,=\,2k-1$

    The sum is: .$\displaystyle S \;=\;\frac{n}{2}\big[2a + (n-1)d\big]$

    . . . . . . . . . $\displaystyle S \;=\;\frac{2k-1}{2}\big[2(k^2-2k+2) + (2k-2)1\big]$

    . . . . . . . . . $\displaystyle S \;=\;\frac{2k-1}{2}\big[2k^2 - 4k+ 4 + 2k - 2\big] $

    . . . . . . . . . $\displaystyle S \;=\;\frac{2k-1}{2}\big[2k^2 - 2k + 2\big]$

    . . . . . . . . . $\displaystyle S \;=\;(2k-1)(k^2 - k + 1)$

    . . . . . . . . . $\displaystyle S \;=\;2k^3 - 2k^2 + 2k - k^2 + k - 1$

    . . . . . . . . . $\displaystyle S \;=\; 2k^3 - 3k^2 + 3k - 1$

    . . . . . . . . . $\displaystyle S \;=\;(k^3 - 3k^2 + 3k - 1) + k^3$

    . . . . . . . . . $\displaystyle S \;=\;(k-1)^3 + k^3$
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    Re: Mathematical Induction

    I'm curious. You replied to your own post from 5 years ago with an almost identical one?

    -Dan
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