Sequence Problem

F

FChan

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Oct 21, 2019
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I have this GP problem and I have the answer but I don’t understand.
Would anyone can help to explain ?


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Have a look at this LINK After that maybe you can explain more about what you do not understand.
Did you setup that summation? Do you understand variance?
It is not clear from the post what help you need.
 
Factor out what is common.
[math](1/4)[1^2(3/4)^0 + 2^2(3/4)^1 + 3^2(3/4)^2 +...] = (1/4)(4/3)[1^2(3/4)^1 + 2^2(3/4)^2 + 3^2(3/4)^3 +...][/math]Does this help you?
 
pka, I can’t understand how it jump to answer 12:
No step is shown.
Please ignore variance, it likes a GP problem.
Thx.
 
Jomo, please show how to come out the answer 12.
It seems like a GP
 
FChan said:
Jomo, please show how to come out the answer 12. It seems like a GP
Click to expand...
It is NOT a GP, because:

\(\displaystyle \frac{a_{n+2}}{a_{n+1}} \ne \frac{a_{n+1}}{a_{n}}\)
 
FChan said:
pka, I can’t understand how it jump to answer 12:
No step is shown.
Please ignore variance, it likes a GP problem.
Click to expand...
FChan, I have absolutely no idea who you think you are. This is a free help site. No one here is a paid helper so none is obligated to follow your instructions. You say to ignore variance when that is the main object of your question.
We have no idea to what data the variance applies. I asked you to tell what you understand about the problem.
If you setup the infinite summation you did a rather sloppy job. That why I showed you a site that you can use to find the sum.
NOW to be perfectly honest, I have absolutely no idea how the \(\displaystyle 4^2\text{ or the }12\) fits into the question.
Chan: this question is about variance on a data set that we have no excess to.
 
FChan said:
I have this GP problem and I have the answer but I don’t understand.
Would anyone can help to explain ?


View attachment 14248
Click to expand...
As I understand it, this is a solution you were given, and you want to understand it. It appears to be finding the variance of a distribution where x = 1, 2, 3, ..., and [MATH]P(x) = (3/4)^{x-1}(1/4)[/MATH]. But you are not asking about that part, only about how to determine the sum

[MATH]\sum_{k=1}^{\infty}k^2(3/4)^{k-1}(1/4) = 28[/MATH],​

in order to fill in the step. (The mean must have been previously determined to be 4, which accounts for the rest of the expression.)

There is probably a specific name for this type of series (it most definitively is not geometric), and a standard method for evaluating it; but since it does indeed have some similarities to geometric series, so try doing what you do to sum such a series! If the sum is S, and r = 3/4, try finding T = S - rS. That won't be something you can immediately sum, but try repeating the same general idea again. You'll get somewhere, and you won't be depending on someone else to give you answers. (I almost got the correct answer, but probably made a little mistake somewhere.)

Now, it would be very helpful if you would tell us about the context of the problem, which might make it easier for us to see what you need to do (and might even let us find an easier way to solve your problem). And it would definitely have helped if you had not called it a "GP problem", confusing everyone. The series (not a GP) is just the one step in the work you showed that you are asking about, not the whole problem.
 
Dr. Peterson, thank you for your patience (your 1st paragraph is exactly my situation). Your reply is really helpful. I totally understand now.

Thanks so much again and again.
 
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