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Mathematics 14 Online
OpenStudy (anonymous):

Given the geometric sequence where a1 = 4 and the common ratio is 3, what is the domain for n? All integers where n ≥ 1 All integers where > 1 All integers where n ≥ 4 All real numbers

OpenStudy (iluvsoccer):

I'm not able to be of assistance, I'm sorry. They can try and help, if not, tag some more people. :) @ParthKohli @Hero

OpenStudy (iluvsoccer):

@iGreen

OpenStudy (bibby):

start by writing out the first few terms of the sequence. 4, 12, 36, ... do you see any restrictions on the lower end?

OpenStudy (anonymous):

what do u mean restrictions?

OpenStudy (bibby):

if the first term of the sequence is 4, it can never go lower than 4.

OpenStudy (anonymous):

okay

OpenStudy (bibby):

it can only increase from there because the common ratio is a positive integer

OpenStudy (anonymous):

so the answer is C?

OpenStudy (bibby):

yeah

OpenStudy (anonymous):

thank you so muchhh can u help me with one more question

OpenStudy (bibby):

yeah, sure

OpenStudy (anonymous):

The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 48 and the common ratio is one fourth. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

OpenStudy (bibby):

gimme a second cleaning somethging

OpenStudy (anonymous):

mmkay

OpenStudy (bibby):

can you write out the sigma?

OpenStudy (anonymous):

i cant on my keyboard

OpenStudy (bibby):

You can use the draw button here

OpenStudy (anonymous):

|dw:1425491157571:dw|

OpenStudy (anonymous):

is this correct?

OpenStudy (bibby):

I don't think so, gimme another minute sorry lol

OpenStudy (anonymous):

its gucci

OpenStudy (bibby):

only frick with hoes that rock dolce and gabanna

OpenStudy (bibby):

gabbana*

OpenStudy (bibby):

anyhow, again start by writing out the first few terms \(48,12,3,\dfrac{3}{4},...\)

OpenStudy (bibby):

given that it's a geometric seq, it's of the form, \(a,ar,ar^2,...,ar^{(n-1)}\)

OpenStudy (anonymous):

This looks hard asf

OpenStudy (bibby):

and so, follow the formula \[\sum_{?}^{?}ar^k\] where a is our first term and r is the common ratio

OpenStudy (bibby):

so \[\sum_{k=0}^{\inf}48*\dfrac{1}{4}^k\]

OpenStudy (bibby):

google the sum formula. i think its on the first page here http://www.purplemath.com/modules/series5.htm

OpenStudy (anonymous):

im sorry im really bad at math and im not picking up what ur putting down

OpenStudy (bibby):

The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 48 and the common ratio is one fourth. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population. the way you write sigma notation for geometric series, is first term times the common ratio raised to the sigmas power think about why this works. if we start counting at 0, anything raised to the 0 power is 1, so we have \(48*\frac{1}{4}^0=48*1=48 \) for our first term

OpenStudy (anonymous):

okay i understand that now.

OpenStudy (bibby):

can you plug in the information we have into this?

OpenStudy (bibby):

wait, whoops

OpenStudy (bibby):

it's just a/(1-r)

OpenStudy (anonymous):

yall be on this for good decade

OpenStudy (bibby):

it feels like 2 or 3 decades. my kids have moved out and my wife left me damn

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