Question

Geometric sum sequence. Hello I'm working on a probability problem about two biased coins, but afterwards I arrive to a geometric series in my work that I am not sure how to calculate since I have not worked with series in a long time. I'll post my work below. will give a medal.

Answer 1

\[\frac{p}{1-p}*\sum_{k=1}^{\infty}(1-p)^{3k}; 0<p<1\] attempt at the solution: know geometric series is \[\sum_{k=0}^{\infty} r^k=\frac{1}{1-r}\] so \[\sum_{k=1}^{\infty}r^{3k}=\frac{1}{1-r^3}-1?\]
so my solution would be \[\frac{p}{1-p}(\frac{1}{1-(1-p)^3}-1)\] However I know this is not the solution, so I am not sure were I am making a mistake or if the formula I am using is correct. Any help is appreciated. Thank you!

Answer 2

0 < p < 1

-1 < p-1 < 0

1 > 1-p > 0

well, since 1-p is less than 1, the series should converge using:\[\sum_{0}^{\infty}r^n=\frac{1}{1-r}\]

Answer 3

are we sure that what we are starting with is valid to the overall solution?

Answer 4

I know that the series converges, however I am not sure how to write what it converges to. Yes, it was a midterm question and I got taken off points for not arriving at the correct solution at the end.

Answer 5

well, lets work this a longer way
\[~~~~~~~S = r^3+r^6+r^9+...+r^{3k}\\
-r^3S = ~~~-r^6-r^9-...-r^{3k}-r^{3k+3}\\
-----------------\\
S=(r^3-r^{3k+3})/(1-r^3)
\]

Answer 6

simplifying we get:
\[S=r^3\frac{1-r^{3k}}{1-r^3}\]
as k to infinity, the top goes to 1 giving us:\[S=\frac{r^3}{1-r^3}\]
seems fair to me

Answer 7

May I know how you arrived to the part after you wrote"---------" Sorry my algebra is a little off and I don't see it.

Answer 8

1S - r^3S = (1-r^3)S

the right side of it is just subtracting all the common terms
and we are left with r^3 - r^3(r^3k)

we want to find S (the Sum) so we divide off the excess.

Answer 9

I am not sure how it is 1S. Wouldn't an r^3 be missing in order to have a complete S value?

Answer 10

I see you added it afterwards

Answer 11

we want to find the sum of a set of terms:

S = r^3 + r^6 + r^9 + ... + r^3k

the historical approach to this is to subtract a modified version
of itself from it to make life simpler: S - r^3 S

r^3 S = r^6 + r^9 + ... + r^3k + r^3k r^3

so now all that we have to do is subtract one from the other

S = r^3 + r^6 + r^9 + ... + r^3k
-r^3 S = -r^6 - r^9 - ... - r^3k - r^3k r^3
^^^^^^^^^^^^^^^^^
all these go to 0

therefore

S - r^3 S = r^3 - r^3k r^3
^^^^^^^
factor the S out


S(1 - r^3) = r^3 - r^3k r^3

and divide off the excess to solve for S, the sum we wanted to know to begin with

Answer 12

as k approaches infinity, r^3k goes to 0 since its less than 1, this leaves us with the formula:

r^3/(1-r^3)

Answer 13

As my solution I now get \[\frac{ p }{ 1-p }[\frac{(1-p)^3}{1-(1-p)^3}-1]\] However when I type in my answer to a site like wolfram alpha to compare my solutions don't match up: http://www.wolframalpha.com/input/?i=sum+from+1+to+infinity+of+p%281-p%29%5E%283k-1%29

http://www.wolframalpha.com/input/?i=p%2F%281-p%29%5B%281-p%29%5E3%2F%5B1-%281-p%29%5E3%5D-1%29%5D%5D
Am I still making a mistake somewhere?

Answer 14

that last -1 is not called for

Answer 15

Why? Isn't it since I start from k=1 to infinity wouldn't I have to subtract the k=0 term?

Answer 16

thats why i worked it the long way.

S starts at r^3 and i simply took it from there to reach the finale

mathing it out avoids trying to work out any memory tricks for adjustments

Answer 17

the mathing process i worked out is correct:

\[S=\frac{r^3}{1-r^3}\]

Answer 18

if we had started with r^3(0) then we would had removed it, but i started at r^3(1)

Answer 19

Oh I think I see it. So in order to work out something that is missing a few terms we subtract the initial terms missing? However I think my answer still does not match with the one on Wolfram Alpha. Is my substitution correct without the -1 at the end or did I make a small mistake still? Sorry for the plethora of questions.

Answer 20

you have a few different wolf links, can you post the 'correct' one for me?

Answer 21

ok

Answer 22

This one is the link to the solution they arrive at:
http://www.wolframalpha.com/input/?i=sum+from+1+to+infinity+of+p%281-p%29%5E%283k-1%29
Afterwards I type in our answer and try to match it with theirs and seeing the variations and graphs to see if they match up. Here is the link to my answer:
http://www.wolframalpha.com/input/?i=p%2F%281-p%29%5B%281-p%29%5E3%2F%5B1-%281-p%29%5E3%5D%29%5D%5D

Answer 23

just a cursory working i get to
\[\frac{p}{1-p}\frac{(1-p)(1-p)(1-p)}{1-(1-p)^3}\]
\[\frac{p(1-p)(1-p)}{1-(1-p)^3}\]
\[\frac{p(p^2+1-2p)}{1-(1-p)^3}\]
\[\frac{p(p^2+1-2p)}{1-(1-3p+3p^2-p^3)}\]
\[\frac{p(p^2+1-2p)}{3p-3p^2+p^3}\]
\[\frac{p(p^2+1-2p)}{p(3-3p+p^2)}\]
\[\frac{p^2-2p+1}{p^2-3p+3}\]

Answer 24

I think it is almost the same solution since you can make the top \[(p-1)^3\] and include at the bottom (p-1) and factor out a negative one I see it now. Thank you so much. Maybe I mistyped it in wolfram alpha. I am not sure though

Answer 25

sum of:
p (1-p)^(3k-1)

the wolf link to this is saying it doesnt understand your input

Answer 26

I'll take a screen shot and attach it

Answer 27

I'll retype it on wolfram alpha first. Sorry my screen shots came out dark gray

Answer 28

your attempt is at trying:\[\frac{p}{1-p}~\frac{(1-p)^3}{\color{red}{1}}=p(1-p)^{3k-1}\]
what happened to the demoninator of our summation?

Answer 29

\[\sum(1-p)^{3k}=\frac{(1-p)^3}{1-(1-p)^3}\]
\[\frac{p}{1-p}\sum(1-p)^{3k}=\frac{p}{1-p}~\frac{(1-p)^3}{1-(1-p)^3}\]
oh wait, i see it:

so you are trying:\[\sum p(1-p)^{3k-1}=\frac{p}{1-p}~\frac{(1-p)^3}{1-(1-p)^3} \]

Answer 30

i say we test it out, let p=.3 :)

Answer 31

Yes. Ok I will do that:)

Answer 32

Yes it matched! Thank you so much. I understand it now :D Have a great day:D

Answer 33

:) youre welcome