Simple Summations Help: Solving 6n+8 with Index i=4

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The discussion revolves around solving the summation of 6n + 8 with an index starting at i=4. The user expresses confusion about how to apply the summation formulas, particularly regarding the separation of terms and the use of constants. Clarification is sought on whether to use the index value of 4 or another constant when calculating the sum. Participants suggest that the user should consider the relationship between the index and the summation limits, specifically how to adjust the sum to start from i=4. The conversation emphasizes the importance of understanding how to modify summation limits to correctly compute the desired sum.
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I forgot how to do summations... I've been googling all day, but I can't find resources good enough for the types of problems I have. If anyone can provide me with some resources, I'd really appreciate it.

Homework Statement


10087
Ʃ6n+8
i=4

I'm just going to call that S

Homework Equations


Ʃ of i = 1 to n is n(n+1)/2

The Attempt at a Solution


I know I can separate the two terms:
S6n + S8

And then pull the constants out?
6n*S1 + 8S1

But can I use 1 as the new constant or do I have to use 4 since that is the index? Do I just multiply 4 by 10087? Or 1 by 10087 and then subtract... something? I don't know, the index of 4 is really throwing me off.
 
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It's not clear what you are trying to sum.

The summation you have is indexed by 'i', but the summand uses 'n'. Is there some relationship between 'i' and 'n' which you have not posted?

If you have a formula for a sum which goes from 'i' to 'n', how would you use that formula if 'i' starts at 4?
Hint: write out the terms of a sum which starts at 1 and goes through say i = 5. How would you modify the sum so that the first term is for i = 4?
 
I know that there's no relationship between i and n, which is why I'm so lost. If I was using i, and it started at 4, would I just subtract the sum of 1 to 3?
 
Of course.
 
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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