- #1

ReubenDS

- 3

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- 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?

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?