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Finding the sum of a series by grouping

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    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


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


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    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


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    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##.

  7. Sep 11, 2016 #6


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    ##\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##
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