Results 1 to 6 of 6

Math Help - Induction and formulating equations

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    2

    Induction and formulating equations

    While browsing, I came across equations for finding the sums of certain series's, the first is (n+n)/2 for "1+2+3...+n", the second is (1/6)n(n+1)(2n+1) for "1+2+3...n".

    I found out how to come up with the first equation through reasoning: n(n+1)/2 is basically the median times n, and since the difference in this series is 1, and the series begins with 1, the median is the same as the mean, so the mean times the number of numbers in the series finds the total.

    Now what I can't figure out is how to formulate the second equation, any ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    2
    Why the 3x multiplier in 3(1+2+...+n)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771
    The shaded area on the right is equal to the total area of the shaded squares on the left (small squares with the same type of shading from various big squares are collected together). Also, in the picture on the right, the blank area to the left of the shaded figure is equal to 1*1 + 2*2 + 3*3 + 4*4 + 5*5: look at it bottom up. The same goes for the blank area right of the shaded figure. Thus, the total area of the rectangle on the right is three times the total area of the squares on the left.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294
    And you already know that 1 + 2 + 3 + \dots + n = \frac{1}{2}n(n+ 1).

    So that means

    3(1^2 + 2^2 + 3^2 + \dots + n^2) = (1 + 2 + 3 + \dots + n)(2n + 1)

     = \frac{1}{2}n(n + 1)(2n + 1).


    Thus

    1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6}n(n + 1)(2n + 1).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by jspstorm View Post
    While browsing, I came across equations for finding the sums of certain series's, the first is (n+n)/2 for "1+2+3...+n", the second is (1/6)n(n+1)(2n+1) for "1+2+3...n".

    I found out how to come up with the first equation through reasoning: n(n+1)/2 is basically the median times n, and since the difference in this series is 1, and the series begins with 1, the median is the same as the mean, so the mean times the number of numbers in the series finds the total.

    Now what I can't figure out is how to formulate the second equation, any ideas?
    You could also create some numerical patterns as follows...

    1^2+2^2+3^2+4^2+5^2+...+n^2=1+4+9+16+25+...+n^2

    =1+(1+3)+(1+3+3+2)+(1+3+3+3+2+2+2)+(1+3+3+3+3+2+2+  2+2+2+2)+....

    =1[n]+3[0+1+2+3+4+5+...+(n-1)]+2[0+0+1+3+6+10+15+21+...]

    Two of these are very straightforward to evaluate.

    1 appears n times.
    3 appears in (n-1) brackets and the sum is a simple arithmetic series of (n-1) terms.

    2 appears in (n-2) brackets and the sum is the sum of triangular numbers
    whose sequence can be seen on the 3rd diagonal from the top of Pascal's triangle.
    The sum of triangular numbers can be seen on the diagonal below that (the 4th from the top).

    1+3=4,\;\;4+6=10,\;\;10+10=20,\;\;20+15=35......

    These sums are \binom{n+2}{3}

    As there are (n-2) terms in our sum, the sum within brackets of the 2's is \binom{n}{3}

    Hence

    \displaystyle\ 1^2+2^2+3^2+...+n^2=n+3\frac{(n-1)(n)}{2}+2\frac{n(n-1)(n-2)}{3!}

    =\displaystyle\ n+3\frac{n(n-1)}{2}+2\frac{n\left(n^2-3n+2\right)}{6}

    =\displaystyle\frac{6n+9n(n-1)+2n\left(n^2-3n+2\right)}{6}

    =\displaystyle\frac{n\left(2n^2-6n+4+9n-9+6\right)}{6}

    =\displaystyle\frac{n\left(2n^2+3n+1\right)}{6}=\f  rac{n(2n+1)(n+1)}{6}
    Attached Thumbnails Attached Thumbnails Induction and formulating equations-sum-squares.jpg  
    Last edited by Archie Meade; September 22nd 2010 at 03:32 AM. Reason: added diagram
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Formulating an expression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 12th 2011, 08:21 AM
  2. [SOLVED] Formulating the BV problem
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: March 24th 2011, 12:41 PM
  3. Formulating a NLP
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 9th 2010, 05:47 PM
  4. Formulating a NLP
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 28th 2010, 02:24 AM
  5. formulating this problem..
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 17th 2009, 01:44 AM

Search Tags


/mathhelpforum @mathhelpforum