How Can Nested Summation Functions Be Simplified or Reversed?

[ReubenDS]

TL;DR

Summation of Summation method
Extracting value for n of a Summation

1. I have come across a few times I would like a more straightforward way to run a summation function on a summation function. I don't have the educational groundwork to know if there is another way to do this or a good technique to simplify these problems.
example:
2*Sum[2x-1*-Sum[Divide[2x-1,2],{x,1,n}],{x,1,n}]
Where 2* summation of all whole integers of the function (2x-1)*-summation of all whole integers of the function (2x-1)/2 where x=1 to n for both summations.

2. It has struck me from time to time that I would like to reverse a summation function to derive an unknown limit for a known resultant of the summation.
example:
(Sum[2x-1,{x,1,n}])=2704
Where the summation of all whole integers of the function 2x-1 for x=1 to n, equals 2704 what is the value of n?

 

[fresh_42]

Could you rephrase this with Latex? Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

It is hard to read in that programming style.

 

[ReubenDS]

Example 1
2 \sum_{x=1}^n 2x-1- \sum{x=1}^n \frac {2x-1} 2

Example 2
\sum_{x=1}^n 2x-1 = 2704

**I do not care for the answers to these example problems

 

[fresh_42]

**I do not care for the answers to these example problems

What do you care for? I have difficulties understanding your question. Do you want to know how to read $\sum_{k=1}^n a_k$ or how to program a sum?

 

[ReubenDS]

I've formatted it in Latex, and I can't explain why it doesn't display correctly.

The question is about proper notations I should be using or techniques to simplify down to my current notation.

 

[fresh_42]

If we want to sum up an expression, say $f(k)$ that depends on a natural number $k$ from some lower value $L$ of $k$ to some upper value $U$ of $k$. That is
$$
f(L)+f(L+1)+f(L+2)+\ldots+f(U-2)+f(U-1)+f(U)=\sum_{k=L}^U f(k)
$$
For example $1+2+3+\ldots+n=\displaystyle{\sum_{k=1}^n k} =\dfrac{n(n+1)}{2}$ or
$$
9+16+25+\ldots+n^2=\sum_{k=3}^n k^2=\left(\sum_{k=1}^n k^2\right)-4-1=\dfrac{n(n+1)(2n+1)}{6}-5
$$

 

[Stephen Tashi]

Example 1
2 \sum_{x=1}^n 2x-1- \sum{x=1}^n \frac {2x-1} 2

What does the notation $\sum x$ mean in this context? What are the limits on that summation?

Example 2
\sum_{x=1}^n 2x-1 = 2704

I'll assume you are asking about the result of $\sum_{x=1}^n P(x)$ where $P(x)$ is a polynomial in $x$ (such as P(x) = 2x -1 ). Such a sum can be expressed as $Q(n)$ where $Q(n)$ is a polynomial in $n$. Apparently you are asking about solving $Q(n) = c$ where $c$ is a specified value. That's the problem of solving for the root(s) of polynomial $Q(n) - c$.

Finding the function $Q(n)$ is covered in the mathematical topic of "The Calculus of Finite Differences".

The topic of finding the roots of polynomials is covered in elementary and advanced courses on algebra.