SOLUTION: If someone could please help me with this problem, it would be greatly appreciated! I have to find the sum of the geometric sequence {{{1/6}}}, {{{1/18}}}, ... {{{1/486}}} T

Algebra ->  Sequences-and-series -> SOLUTION: If someone could please help me with this problem, it would be greatly appreciated! I have to find the sum of the geometric sequence {{{1/6}}}, {{{1/18}}}, ... {{{1/486}}} T      Log On


   

Question 266760: If someone could please help me with this problem, it would be greatly appreciated!
I have to find the sum of the geometric sequence
1%2F6, 1%2F18, ... 1%2F486
This is what I have so far:
r=(1/3)
+%281%2F486%29+=+%281%2F6%29%2A%281%2F3%29%5En-1+
(it's supposed to be (n-1))
from this I got n=5

+%281%2F6%29%2A%281-%281%2F3%29%5E5%29+%2F+%281-%281%2F3%29%29+
+%281%2F6%29%2A%281-%281%2F243%29%29+%2F+%282%2F3%29+
As elementary as this is, I need help finding the sum and understanding the process(es) of how to work this equation when fractions are in the mix. (That is, unless I've made mistakes prior to getting to this point.)
Found 2 solutions by Edwin McCravy, stanbon:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
If someone could please help me with this problem, it would be greatly appreciated!
I have to find the sum of the geometric sequence
1%2F6, 1%2F18, ... 1%2F486
This is what I have so far:
r=1%2F3
+1%2F486+=+%281%2F6%29%2A%281%2F3%29%5E%28n-1%29+
(it's supposed to be (n-1))
from this I got n=5

+%281%2F6%29%2A%281-%281%2F3%29%5E5%29+%2F+%281-%281%2F3%29%29+
+%281%2F6%29%2A%281-%281%2F243%29%29+%2F+%282%2F3%29+
As elementary as this is, I need help finding the sum and understanding the process(es) of how to work this equation when fractions are in the mix. (That is, unless I've made mistakes prior to getting to this point.)

You're doing fine.  Now change the division by 2%2F3
to a multiplication by 3%2F2  [Invert and multiply]

+%281%2F6%29%2A%281-1%2F243%29%2A+%283%2F2%29+

change the 1 to 243%2F243

+%281%2F6%29%2A%28243%2F243-1%2F243%29%2A+%283%2F2%29+

+%281%2F6%29%2A%28242%2F243%29%2A+%283%2F2%29+

Now you should be able to handle that with some
cancelling.  You should end up with 

121%2F486

Edwin

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I have to find the sum of the geometric sequence
+%281%2F6%29%2C+%281%2F18%29%2C+...+%281%2F486%29+
This is what I have so far:
r=(1/3)
+%281%2F486%29+=+%281%2F6%29%2A%281%2F3%29%5E%28n-1%29+
(it's supposed to be (n-1))
from this I got n=5

+%281%2F6%29%2A%281-%281%2F3%29%5E5%29+%2F+%281-%281%2F3%29%29+
[(1/6)*(1-(1/243))] / (2/3)
---
= [(1/6) -(1/1458)]/(2/3)
---
= [(243/1458)-(1/1458)]/(2/3)
---
= [242/1458]/(2/3)
---
= (3/2)*(242/1458)
= 121/486
=============================
Cheers,
Stan H.
==================================


As elementary as this is, I need help finding the sum and understanding the process(es) of how to work this equation when fractions are in the mix. (That is, unless I've made mistakes prior to getting to this point.)