Solving Series Summation Problem: Start & How-To

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SUMMARY

The discussion focuses on solving a series summation problem involving the cubic series. Specifically, it addresses the summation of \( r^3 \) from \( r=1 \) to \( n \) and from \( r=n+1 \) to \( 2n \). The established results are \( \sum_{r=1}^{n} r^3 = \frac{n^2}{4}(n+1)^2 \) and \( \sum_{r=n+1}^{2n} r^3 = \frac{n^2}{4}(3n+1)(5n+3) \). The recommended approach to prove these statements is through mathematical induction, starting with the base case of \( n=1 \).

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Students, educators, and mathematicians who are working on series summation problems or teaching mathematical induction techniques.

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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