Results 1 to 3 of 3

Math Help - Double Summation (multiplying)

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    1

    Double Summation (multiplying)



    Is there some kind of formula/shortcut for solving something like this? I could write it all out, but seems unpractical.

    What I'm currently using:
    [ 50(50+1)/2 - ( i - 1)(i -1 +1)/2 ] * i
    Just fill out that 50 times .

    I made a program to solve it and it gave out: 834275.
    Last edited by krazyxazn; February 28th 2010 at 01:54 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, krazyxazn!

    Your answer is correct!

    We're expected to evaluate it without cranking out the sums,
    but it still takes some knowledge and quite a bit of algebra.


    S \;=\;\sum^{50}_{i=1} \sum^{50}_{j=i} ij

    The inner sum is:

    . . \sum^{50}_{j=1} ij \;=\;i(i) + i(i+1) + i(i+2) + i(i+3) + \hdots + i(50)

    . . . . . . =\;i \underbrace{\bigg[i + (i+1) + (i+2) + (i+3) + \hdots + 50\bigg]}_{T = \text{ sum of integers from }i\text{ to } 50}

    . . T \;=\;\frac{50\cdot51}{2} - \frac{(i-1)i}{2} \;=\;\frac{2550+i-i^2}{2}

    Hence, the inner sum is: . \sum^{50}_{j=i} ij \;=\;\left(\frac{2550+i - i^2}{2}\right) i \;=\;\frac{1}{2}(2550i + i^2 - i^3)


    The entire sum is: . S \;=\;\sum^{50}_{i=1}\frac{1}{2}(2550i + i^2  - i^3)

    And we have: . S \;=\;1275\sum^{50}_{i=1} i \;+\;\frac{1}{2}\sum^{50}_{i=1} i^2 \;-\;\frac{1}{2}\sum^{50}_{i=1} i^3


    We are expected to know the formulas for these three sums. . **

    . . S \;=\;1275\cdot\frac{50\cdot51}{2} \;+\;\frac{1}{2}\cdot\frac{50(51)(101)}{6} \;-\; \frac{1}{2}\cdot\left(\frac{50\cdot51}{2}\right)^2


    Therefore: . \boxed{S \;=\;834,\!275}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    ** . Formulas


    . . \sum^n_{i=1}i \;\;\,=\quad 1+2+3+\hdots + n \quad =\;\; \frac{n(n+1)}{2}

    . . \sum^n_{i=1} i^2 \;=\; 1^2+2^2+3^2+\hdots + n^2 \;=\; \frac{n(n+1)(2n+1)}{6}

    . . \sum^n_{i=1} i^3 \;=\; 1^3 + 2^3 + 3^3 + \hdots + n^3 \;=\; \left[\frac{n(n+1)}{2}\right]^2

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by krazyxazn View Post


    Is there some kind of formula/shortcut for solving something like this? I could write it all out, but seems unpractical.

    What I'm currently using:
    [ 50(50+1)/2 - ( i - 1)(i -1 +1)/2 ] * i
    Just fill out that 50 times .

    I made a program to solve it and it gave out: 834275.
    Alternatively, \sum_{i=1}^{50}\sum_{j=i}^{50}ij=\sum_{i=1}^{50}\l  eft\{\sum_{j=1}^{50}ij-\sum_{j=1}^{i-1}ji\right\}. The inner sum is i\left(\cdot\sum_{j=1}^{50}j-\sum_{j=1}^{i}j\right)=i\left(\frac{50\cdot 51}{2}-\frac{i\cdot (i+1)}{2}\right)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expand Double Summation
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: September 6th 2011, 05:54 PM
  2. Double Summation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 20th 2009, 07:11 AM
  3. Double summation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: August 23rd 2009, 06:09 PM
  4. double summation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 28th 2009, 09:15 PM
  5. double summation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 30th 2009, 12:11 PM

Search Tags


/mathhelpforum @mathhelpforum