Prove that these two series are equal

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Homework Help Overview

The discussion revolves around proving the equality of two series, specifically comparing the left-hand side (LHS) and right-hand side (RHS) expressions. Participants are examining the validity of the series as presented in a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are analyzing the series by calculating the LHS and RHS, noting discrepancies in their results. Some question whether there is a potential error in the textbook, while others explore how to modify the original question to achieve equality. There is also mention of a substitution method to re-write the LHS.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning the correctness of the original problem statement. Some guidance has been offered regarding a substitution method, but there is no consensus on how to resolve the perceived error or modify the question effectively.

Contextual Notes

Participants are operating under the assumption that there may be an error in the textbook, particularly concerning the RHS of the equation. The original problem's setup and the nature of the series are under scrutiny.

chwala
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Homework Statement
The question is attached as it appears on the textbook.
Relevant Equations
sum of series
1610448349811.png

i looked at this and it was not making any sense at all, could it be a textbook error or i am missing something here; note that,
lhs gives us, ##4,6,8,10,12,14##
rhs gives us, ##8,11,14,17,20,23##
 
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[tex]LHS=\sum_{k=1}^6 (2(k+2)-2)=\sum_{k=1}^6 (2k+2)[/tex]
[tex]RHS=\sum_{k=1}^6 (3k+5)[/tex]
So
[tex]RHS-LHS=\sum_{k=1}^6 (k+3)= \frac{6*7}{2}+3*6=39\neq 0[/tex]
 
anuttarasammyak said:
[tex]LHS=\sum_{k=1}^6 (2(k+2)-2)=\sum_{k=1}^6 (2k+2)[/tex]
[tex]RHS=\sum_{k=1}^6 (3k+5)[/tex]
So
[tex]RHS-LHS=\sum_{k=1}^6 (k+3)= \frac{6*7}{2}+3*6=39\neq 0[/tex]

therefore there is an error on the textbook...
 
Ok, then what could have been the right question?...,assuming that an error was made. How do we modify the original question in order for the proof to hold (i am assuming that the error made was on the right hand side of the equation)...
 
You can try and work that out yourself! Use the substitution ##u = k-2## to re-write the sum on the LHS
 
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etotheipi said:
You can try and work that out yourself! Use the substitution ##u = k-2## to re-write the sum on the LHS

has that not been done in post number ##2##?
 
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