Struggling with Calculating a Series? Get Help Here!

[alias25]

sorry isn't a physics question more to do with maths.

is it ok to post it here? I can't see to delete it, maybe some one will move it for me? to the maths section. thanksL:)

I have to calculate:

sum of (from r=1 to n): 1/3 (r^3 - (r-1)^3 -1)

so far I've done:

let n = 3

so

sum = 1/3 (n^3 - (n-1)^3 -1)

+1/3 ( (n-1)^3 - (n-2)^3 - 1)

+1/3 ( (n-2)^3 - (n-3)^3 -1)


looking at ^

(n-3)^3 = 0

the 1/3 can be factorised out, the -1's sums to -n

so: sum = 1/3 (n^3 -...-n) (1)

this inbetween thing I am finding tricky

i noticed theres...-2(n-1)^3 - 2(n-2)^3...etc, when n = any number.

i can put that as

sum(from r=1 to n-1) of: (n-r)^3

so into (1):

1/3 (n^3 -2(sum from r=1 to n-1 ofn-r)^3) -n)

^but i don't think that's what they are looking for.

any help will be appriciated,thank you.

 

[neutrino]

Is this what you need to find $$\sum_{r=1}^{n}\left(\frac{1}{3}\left({r^3 - (r-1)^3 - 1}\right)\right)$$?

If so, why not simplify the term that is to be summed, i.e., $\frac{1}{3}\left(r^3 - (r-1)^3 - 1\right)$ and then sum it from 1 to n?

 

[kyle8921]

I understand that mathematics is a fundamental tool for solving problems in many scientific fields, including physics. While this may not be a physics-specific question, it is still a valid question that requires mathematical skills to solve. It is perfectly acceptable to post it here, and I'm sure someone from the community will be able to help you with your calculations. However, if you feel that your question would be better suited for the math section, you can always ask a moderator to move it for you. In the meantime, I would suggest breaking down the problem into smaller parts and using algebraic manipulations to simplify the expression. Good luck!