# Finding the sum of a series by grouping

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1. Sep 3, 2016

### sooyong94

1. The problem statement, all variables and given/known data

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. Sep 3, 2016

### Math_QED

You can use mathematical induction.

3. Sep 3, 2016

### SammyS

Staff Emeritus
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.

4. Sep 4, 2016

### thelema418

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
5. Sep 4, 2016

### ehild

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

6. Sep 11, 2016

### ehild

$\sum_1^n{[(3k-2)^3 +(3k-1)^3 +(3k)^3]} =\sum_1^{3n}{k^3}$
and $\sum_1^n{(3k)^3}=3^3\sum_1^n{k^3}$
Then use the theorem $\sum_1^n k^3=\left(\sum_1^n k \right)^2$