Find Sum of Arithmetic Series Sn: Σ 200 r=5 5r-2

AI Thread Summary
The discussion revolves around calculating the sum of the arithmetic series defined by the expression Σ from r=5 to 200 of (5r - 2). The initial calculation yielded an incorrect sum of approximately 101400, while the correct value verified by a calculator is 100058. Participants emphasize the importance of clarity in mathematical notation, suggesting that parentheses can significantly alter the interpretation of the sum. There is uncertainty about the original intent of the expression, prompting a request for confirmation from the original poster. The conversation highlights the need for precise notation in mathematical discussions to avoid confusion.
Nubcake
Messages
35
Reaction score
0
Find the sum of \Sigma 200 r=5 5r-2

Sn = n/2 [2a + (n-1)d ]I used S 200 and I got about 101400 but then when I verified on my calculator it was 100058, my calculator has the sigma notation for working out the sum of , how do you get 100058?
 
Physics news on Phys.org
Nubcake said:
Find the sum of \Sigma 200 r=5 5r-2

Sn = n/2 [2a + (n-1)d ]


I used S 200 and I got about 101400 but then when I verified on my calculator it was 100058, my calculator has the sigma notation for working out the sum of , how do you get 100058?

It's difficult to tell what your sum is. Is this it?

Edit: add parentheses for clarity.
$$ \sum_{r = 5}^{200} \left(5r - 2\right)$$

Writing this in LaTeX is not difficult. Click what I wrote to see how I did it.
 
Last edited:
Mark44 said:
It's difficult to tell what your sum is. Is this it?
$$ \sum_{r = 5}^{200} 5r - 2$$

Writing this in LaTeX is not difficult. Click what I wrote to see how I did it.

What you wrote coul be reasonably interpreted as
\left( \sum_{r=5}^{200} 5 r \right) -2; it might be that the OP wants
\sum_{r=5}^{200} (5 r - 2),
but who knows?

RGV
 
Ray Vickson said:
What you wrote coul be reasonably interpreted as
\left( \sum_{r=5}^{200} 5 r \right) -2; it might be that the OP wants
\sum_{r=5}^{200} (5 r - 2),
but who knows?
That's a good point. When I wrote that I was going by a sort of quasi-convention that the summation symbol acts like a giant left parenthesis, and there is an implied right paren to the right of the last symbol. If you want to indicate that something is not part of the sum you put parentheses into suit.

Since what I wrote could be interpreted in a couple of ways, I went back and put in some parentheses.

I wasn't sure what the OP intended, so now I guess we wait until he/she comes back to confirm that our guesses are correct.
 
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
Back
Top