Prove that these two series are equal

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SUMMARY

The discussion centers around proving the equality of two series, specifically the left-hand side (LHS) defined as LHS = ∑_{k=1}^6 (2(k+2)-2) and the right-hand side (RHS) defined as RHS = ∑_{k=1}^6 (3k+5). The calculations reveal that RHS - LHS equals 39, indicating a discrepancy and suggesting a potential error in the textbook. Participants propose modifying the original question and using the substitution u = k-2 to re-write the LHS for further analysis.

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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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LHS=\sum_{k=1}^6 (2(k+2)-2)=\sum_{k=1}^6 (2k+2)
RHS=\sum_{k=1}^6 (3k+5)
So
RHS-LHS=\sum_{k=1}^6 (k+3)= \frac{6*7}{2}+3*6=39\neq 0
 
anuttarasammyak said:
LHS=\sum_{k=1}^6 (2(k+2)-2)=\sum_{k=1}^6 (2k+2)
RHS=\sum_{k=1}^6 (3k+5)
So
RHS-LHS=\sum_{k=1}^6 (k+3)= \frac{6*7}{2}+3*6=39\neq 0

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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