# Some Summation inquiries

• I
• ReubenDS
ReubenDS
TL;DR Summary
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?

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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

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**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.

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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
$$9+16+25+\ldots+n^2=\sum_{k=3}^n k^2=\left(\sum_{k=1}^n k^2\right)-4-1=\dfrac{n(n+1)(2n+1)}{6}-5$$

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.