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
I'm not able to be of assistance, I'm sorry. They can try and help, if not, tag some more people. :) @ParthKohli @Hero
@iGreen
start by writing out the first few terms of the sequence. 4, 12, 36, ... do you see any restrictions on the lower end?
what do u mean restrictions?
if the first term of the sequence is 4, it can never go lower than 4.
okay
it can only increase from there because the common ratio is a positive integer
so the answer is C?
yeah
thank you so muchhh can u help me with one more question
yeah, sure
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.
gimme a second cleaning somethging
mmkay
can you write out the sigma?
i cant on my keyboard
You can use the draw button here
|dw:1425491157571:dw|
is this correct?
I don't think so, gimme another minute sorry lol
its gucci
only frick with hoes that rock dolce and gabanna
gabbana*
anyhow, again start by writing out the first few terms \(48,12,3,\dfrac{3}{4},...\)
given that it's a geometric seq, it's of the form, \(a,ar,ar^2,...,ar^{(n-1)}\)
This looks hard asf
and so, follow the formula \[\sum_{?}^{?}ar^k\] where a is our first term and r is the common ratio
so \[\sum_{k=0}^{\inf}48*\dfrac{1}{4}^k\]
google the sum formula. i think its on the first page here http://www.purplemath.com/modules/series5.htm
im sorry im really bad at math and im not picking up what ur putting down
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
okay i understand that now.
can you plug in the information we have into this?
wait, whoops
it's just a/(1-r)
yall be on this for good decade
it feels like 2 or 3 decades. my kids have moved out and my wife left me damn
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