Changing the Limits of Summation

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The discussion focuses on changing the limits of summation in mathematical expressions. The user seeks clarification on how to derive a shortcut for the summation from r=m to n, specifically how to change the index from r to s. The explanation involves redefining the index and demonstrating that the sums remain equivalent despite the change in variable names. The user requests further clarification and examples to solidify their understanding of the concept. Overall, the discussion emphasizes the flexibility of index notation in summation without altering the outcome.
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Homework Statement



Dear Mentors and PF helpers,

Here's my question, I see these on my textbook but couldn't really understand how they derived this short cut.
Please show me how they got to these. Thank you for your time.

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Homework Equations



These is what I understand from now.

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The Attempt at a Solution

 
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So we have the sum \displaystyle \sum_{r=m}^n T_r. We may change the index of summation from r to s=r+k. Then because r starts from m, s should start from m+k and because r ends at n, s should end at n+k. This means that we changed our summation to \displaystyle \sum_{s=m+k}^{n+k}T_{s-k}. But you should note that this sum doesn't depend on s as the first sum didn't depend on r. r and s are just dummy indices because after you do the sums, there'll remain no sign of them. So I can safely rename s to r. But because the sums are just numbers and because the changes I did to the first sum doesn't change the result, I'll have:
\displaystyle \sum_{r=m}^n T_r=\sum_{r=m+k}^{n+k}T_{r-k}.
 
Dear Shyan,

Do you mind giving me a better view with these words, can give your explanation with examples to explain these.

Thank you for your time
 
\displaystyle \sum_{r=m}^n T_r=T_m+T_{m+1}+T_{m+2}+\dots+T_{n-2}+T_{n-1}+T_n

\displaystyle \sum_{r=m+k}^{n+k} T_{r-k}=T_{m+k-k}+T_{m+k+1-k}+T_{m+k+2-k}+\dots+T_{n+k-2-k}+T_{n+k-1-k}+T_{n+k-k}=\\ T_m+T_{m+1}+T_{m+2}+\dots+T_{n-2}+T_{n-1}+T_n

Is it clear enough?
 
Thank you very much Shyan.;)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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