1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the sum of a series by grouping

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-9-3_16-57-5.png

    2. Relevant equations
    Summation

    3. The attempt at a solution
    I know I could have simplified (3n-2)^3 +(3n-1)^3 -(3n)^3 and put the formulas in but I wonder is there any other method (I was thinking about grouping the terms, but to no avail) to work this out.
     
  2. jcsd
  3. Sep 3, 2016 #2

    Math_QED

    User Avatar
    Homework Helper

    You can use mathematical induction.
     
  4. Sep 3, 2016 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Are you given a set of instructions for some group of problems?

    As @Math_QED mention, it does look like a problem typically solved by using induction. However, if you have been given formulas for summing various powers of n, in particular n3, then you can do this using such a formula.
     
  5. Sep 4, 2016 #4
    Hey @sooyong94,

    In your post, I do not understand what you mean by "could have simplified ##(3n-2)^3 +(3n-1)^3 -(3n)^3##. " Could you explain how this simplification method works to confirm the validity of the given equation (Problem Statement)?

    If you have memorized summation laws (or have access to a table), then regrouping is always a viable way to try on these problems. Regrouping is just the Associative Property of Addition.
     
    Last edited by a moderator: Sep 4, 2016
  6. Sep 4, 2016 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    you can write the terms as ##[(3n-2)^3 +(3n-1)^3 +(3n)^3] - [2(3n)^3]## and sum the terms in the brackets separately. What do you get if you write out the sum of the first brackets?
    Then use the theorem ##\sum_1^n k^3=\left(\sum_1^n k \right)^2##.

    https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes
     
  7. Sep 11, 2016 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    ##\sum_1^n{[(3k-2)^3 +(3k-1)^3 +(3k)^3]} =\sum_1^{3n}{k^3}##
    and ##\sum_1^n{(3k)^3}=3^3\sum_1^n{k^3}##
    Then use the theorem ##\sum_1^n k^3=\left(\sum_1^n k \right)^2##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding the sum of a series by grouping
Loading...