Why Does Geometric Sequence General Term Formula Use n, Not n-1?

AI Thread Summary
The discussion clarifies why the general term formula for a geometric sequence uses n instead of n-1. When the sequence starts at zero, using n allows for accurate representation of the terms without resulting in negative indices. If the formula were raised to n-1, it would incorrectly imply a term of -1 for the first element. The participants agree that the choice of n aligns with the starting point of the sequence. Understanding this distinction is essential for correctly applying geometric sequences in mathematical contexts.
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do you have an explanation to why sometimes the general term formula for geometric sequence is not raised to n-1 but to simply n?
 
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because you could have \sum_{n=0}^{\infty} ar^{n} or \sum_{n=1}^{\infty} ar^{n-1}

Or \sum_{n=k}^{\infty} ar^{n-k}
 
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oh ok...so from what I understood, if the sequence starts at zero then the formula raised to n because if it was n-1 then it will read -1, right?
 
that is correct
 
all right thank you!
 
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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