Series notation and commutativity

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Homework Help Overview

The discussion revolves around understanding series notation and transformations involving summation. Participants are examining the expression involving a summation and its manipulation, particularly focusing on the term -3n and its derivation from the sum of constants.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand why the transformation results in -3n instead of simply -3. Some participants provide insights into the definition of multiplication and how it relates to summation.

Discussion Status

The discussion is active, with participants engaging in clarifying the reasoning behind the transformation of the series notation. There are multiple interpretations being explored regarding the manipulation of constants in summation.

Contextual Notes

Participants are working within the constraints of homework rules, focusing on the definitions and properties of summation without providing complete solutions.

vorophobe
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Homework Statement


I'm trying to wrap my head around series notation, but I'm finding some of the transformations hard to grasp. For example, this one:

Homework Equations


\Sigma^{n}_{i=1} (2a_{i}-3) = 2\Sigma^{n}_{i=1}a{_i}(-3n)

The Attempt at a Solution


In the above expression, I don't understand how you end up with -3n on the RHS. Why can't it just be left as -3?
 
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Actually, that is the definition of multiplication!

3n = \overbrace{3 + 3 + ... + 3}^{n \ times} \ = \ \sum_{i=1}^n3
 
vorophobe said:

Homework Statement


I'm trying to wrap my head around series notation, but I'm finding some of the transformations hard to grasp. For example, this one:

Homework Equations


\Sigma^{n}_{i=1} (2a_{i}-3) = 2\Sigma^{n}_{i=1}a{_i}(-3n)

The Attempt at a Solution


In the above expression, I don't understand how you end up with -3n on the RHS. Why can't it just be left as -3?

##\sum_{i = 1}^n (2a_i - 3) = \sum_{i = 1}^n 2a_i - \sum_{i = 1}^n 3##

The last sum is 3 + 3 + 3 + ... + 3, a sum with n terms.
 
Great replies thank you both!
 

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