Results 1 to 6 of 6

Math Help - Mathematical Induction

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    13

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2007
    Posts
    176
    here's my working
    Attached Thumbnails Attached Thumbnails Mathematical Induction-out0071.jpeg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    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: \forall n \in \mathbb{N}

    \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; February 22nd 2008 at 02:11 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, BACONATOR!

    Find the general statement is the biggest task . . .


    18. Consider the following four equations:

    \begin{array}{cccccc}<br />
1) & 1 & = & 1 & = & 0^3+1^3\\<br />
2) & 2+3+4 &= & 1+8 & = & 1^3+2^3\\<br />
3) & 5+6+7+8+9 & = & 8+27 &=& 2^3+3^3\\<br />
4) & 10+11+12+13+14+15+16 & = & 27+64 & = & 3^3+4^3<br />
\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 n: sum of 2n-1 consecutive integers up to n^2, equals (n-1)^3 + n^3


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

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

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

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

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627

    Re: Mathematical Induction

    Hello, BACONATOR!

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

    \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}

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

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

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

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

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

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

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

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

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

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

    . . . . . . . . . S \;=\;(k-1)^3 + k^3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1

    Re: Mathematical Induction

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mathematical Induction
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: August 31st 2010, 03:31 PM
  2. Mathematical induction
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: August 30th 2010, 05:54 AM
  3. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  4. mathematical induction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 13th 2009, 05:29 PM
  5. Mathematical Induction
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 18th 2009, 08:35 AM

Search Tags


/mathhelpforum @mathhelpforum