Solving Series Summation Problem: Start & How-To

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    Series Summation
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Homework Help Overview

The discussion revolves around a series summation problem involving cubic terms, specifically the summation of r^3 from different limits. The original poster is seeking guidance on how to approach the problem and the appropriate starting point for the summation.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the original poster's question regarding the summation limits and mathematical notation. There are suggestions to prove the statements by induction and to manipulate the sums through subtraction and factoring.

Discussion Status

The discussion is ongoing, with some participants providing guidance on potential methods to approach the problem, while others are seeking clarification on the original question. There is no explicit consensus yet on the best way to proceed.

Contextual Notes

Participants are working with a textbook problem rather than a traditional homework assignment, which may influence the nature of the discussion and the expectations for the solution approach.

tykescar
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It isn't homework, it's in a textbook and I'm having trouble with it.

When r=1, summing to n the series of r^3 = (n^2)/4 (n+1)^2

Show that when r = (n+1), summing to 2n = (n^2)/4 (3n+1)(5n+3)

What order do you start the summation, and how do I begin?
 
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I'm sorry, but I just don't understand your question :frown: Can you please use mathematical notation, or explain it better? I know it is supposed to be about sums, but I really can't see what sums...
 
\sum_{r=1}^{n}r^3=\frac{n^2}{4}\right\left(n+1)\right^2

show that

\sum_{r=n+1}^{2n}r^3=\frac{n^2}{4}\right\left(3n+1)\right\left(5n+3)\right

I don't understand how to start
 
Last edited:
Aah, thanks for TeXing it, I understand now!

Let's first do the first sum. You should prove such a statements by induction. So, first prove the case n=1. Secondly, assume that n=k has been shown, and prove the case n=k+1.
 
hi tykescar! :wink:

just do the sum from 1 to 2n, subtract from it the sum from 1 to n, and do a bit of factoring …

what do you get? :smile:
 

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