Analytic Geometry: closed form for 3/5 + 3/25 + ... + 3/5^10
- Thread starter thinker86
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D
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In another post you said you have finished pre-calc and you are good at it.
This problem should be an easy one with those skills.
What is the sum of
\(\displaystyle S\, = \, a \, + \, ar^1 \, + \, ar^2 \, + \, ar^3 \, + ..... \, ar^{(n-1)} \, + \, ar^n\)
In this case
\(\displaystyle a\, = \, \frac{3}{5}\)
and
\(\displaystyle r\, = \, \frac{1}{5}\)
tell us what did you find.
This problem should be an easy one with those skills.
What is the sum of
\(\displaystyle S\, = \, a \, + \, ar^1 \, + \, ar^2 \, + \, ar^3 \, + ..... \, ar^{(n-1)} \, + \, ar^n\)
In this case
\(\displaystyle a\, = \, \frac{3}{5}\)
and
\(\displaystyle r\, = \, \frac{1}{5}\)
tell us what did you find.
Oh okay I think I got it. I did do well in pre-calc but there are a few things I don't remember. I think I got the correct answer.
I worked it out using the formula given to me from above. The problem was I mixed it up with another formula, that's why it wasn't working. The answer that I got was:
(3/4)(1-1/5^10)
Thanks for your help!
I worked it out using the formula given to me from above. The problem was I mixed it up with another formula, that's why it wasn't working. The answer that I got was:
(3/4)(1-1/5^10)
Thanks for your help!
thinker86 said:
Your answer has "boggled" me...I don't understand how you came up with a 3/4 being involved in this problem.
I'd look at it this way:
first term = 3*(1/5)
second term = 3*(1/5)*(1/5)
third term = 3*(1/5)*(1/5)*(1/5)
etc......
now...looks to me like the nth term (that closed form thing you're looking for) ought to be
3 * (1/5)^n where n is the number of the term of interest.
D
Deleted member 4993
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MrsPi,
thinker86 was looking for the sum of the series - not the 'n' th term of the sequence.
In that case, the expression of the sum is correct. The term 3/4 comes from (3/5)/[1- (1/5)]
thinker86 was looking for the sum of the series - not the 'n' th term of the sequence.
In that case, the expression of the sum is correct. The term 3/4 comes from (3/5)/[1- (1/5)]
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The purpose of the placement test for which you are "cramming" (in a mis-guided attempt to "fool" the test) is to figure out what you remember, what you need to review, and where you should most-economically be started in the sequence of math courses. Yes, you may be bored for the first few weeks of the pre-calc course in which you clearly should begin, but that's a good thing. If you start the first day having no idea what is going on, you will flunk, probably repeatedly.thinker86 said:
I've seen this way too many times. Trust me: It is much more expensive to take Calc I six times before barely passing it, Calc II three times, giving up on Diff-EQ, and then changing majors (so a dozen or so alreadly-completed courses become wasted time and money), than to simply accept what the placement test tells you, and take the one or two pre-calc courses first. The placement test is there for a reason; let it do its job.
Eliz.