Summation Problem: Find Sum of 3r - 2r to n Terms

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The discussion centers on finding the sum of the series defined by the rth term 3r - 2r. Initial attempts involved logarithmic manipulation but led to confusion regarding the properties of logarithms. A suggestion was made to split the series into two geometric series for easier calculation. The correct answer is identified as 3n+1/2 - 2n+1 + 1/2. Participants emphasize the importance of reviewing logarithmic properties to avoid similar mistakes in future calculations.
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Homework Statement


Find the sum to n terms of the series whose rth term is 3r - 2r

Homework Equations


The Attempt at a Solution



So I tried this

rlog3 - rlog2 = n(n+1)/2 log3 - n(n+1)/2 log2

then I realized this was kind of useless, the only thing I could get from this is

n(n+1)/2 log (3/2) , then I don't know what to do with it.

the answer in the book is 3n+1/2 -2n+1 + 1/2
 
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lionely said:

Homework Statement


Find the sum to n terms of the series whose rth term is 3r - 2r




Homework Equations





The Attempt at a Solution



So I tried this

rlog3 - rlog2 = n(n+1)/2 log3 - n(n+1)/2 log2
You have a mistake right off the bat. log(A - B) ≠ logA - logB.
Split your series into two - each part is a geometric series.
lionely said:
then I realized this was kind of useless, the only thing I could get from this is

n(n+1)/2 log (3/2) , then I don't know what to do with it.

the answer in the book is 3n+1/2 -2n+1 + 1/2
 
I see it now thanks.. I keep making those darned mistakes with logs!
 
It would be worth your while to review the properties of logs. They might be listed in an appendix of the book you're using. If not, get a precalc book or put in some time in an online tutorial such as khanacademy or the like.
 
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