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OpenStudy (anonymous):

Easy Alg 2 Question :)

OpenStudy (anonymous):

OpenStudy (xapproachesinfinity):

Do you now the sum for a geometric series?

OpenStudy (anonymous):

No @xapproachesinfinity

OpenStudy (xapproachesinfinity):

okay! \[\sum_{n=0}^{\infty} a1r^n\]

OpenStudy (xapproachesinfinity):

that the sum of the geo series where a1 is the first term r is the common ratio now do that for this series!

OpenStudy (anonymous):

@xapproachesinfinity okay so this crosses out A and B, so now I have C and D left

OpenStudy (xapproachesinfinity):

Correct!

OpenStudy (xapproachesinfinity):

you have to take the limit of that sum to determine if it is convergent and do the sum of if it is divergent

OpenStudy (anonymous):

@xapproachesinfinity how do you know the limit of the sum?

OpenStudy (xapproachesinfinity):

limit of the sum is sum of limit so you only need to take the limit of the that expression a1r^n

OpenStudy (anonymous):

@xapproachesinfinity so the answer is D, right?

OpenStudy (xapproachesinfinity):

why?

OpenStudy (anonymous):

@xapproachesinfinity because it diverges when you take the sum of the limit?

OpenStudy (xapproachesinfinity):

Not really what's the limit?

OpenStudy (anonymous):

188?

OpenStudy (xapproachesinfinity):

how did you compute that! i have no idea xD i think the limit is zero?

OpenStudy (anonymous):

UGH I'm so confused lol

OpenStudy (anonymous):

I thought you said to get the limit you solve a1r^n

OpenStudy (anonymous):

and that would be 940(1/5)^1, so i got 188 lol

OpenStudy (xapproachesinfinity):

No

OpenStudy (xapproachesinfinity):

\(\large \it\color{maroon}{ \sum_{i}^{infinity}940(\frac{1}{5})^i }\) \(\large \it\color{maroon}{ \lim_{i \rightarrow \infty}940(\frac{1}{5})^i}\)

OpenStudy (xapproachesinfinity):

that's the limit i told you to find!

OpenStudy (anonymous):

So if the limit goes to infinite it can't have a number, therefore it's diverging?

OpenStudy (anonymous):

@xapproachesinfinity

OpenStudy (xapproachesinfinity):

No what's the limit of that power thing?

OpenStudy (xapproachesinfinity):

it is not going to infinity!!

OpenStudy (anonymous):

@xapproachesinfinity 0?

OpenStudy (xapproachesinfinity):

correct the limit is zero! so the series is convergent actually there a rule that says if r is less than 1 the series should convergent

OpenStudy (anonymous):

okay, so the answer is C? :)

OpenStudy (xapproachesinfinity):

now you need to find the sum

OpenStudy (anonymous):

well if it's convergent then i already know the answer can't be D, and since we eliminated A and B.... :)

OpenStudy (xapproachesinfinity):

well it is the answer C, but can we make sure! i mean we are sure since there are no other options but can we do it and find the sum

OpenStudy (anonymous):

sure

OpenStudy (xapproachesinfinity):

well let's say for now the question is changed and asked to do the sum? what would you do

OpenStudy (anonymous):

also after can you help me with one more?

OpenStudy (anonymous):

you would substitute i into the power and solve it through

OpenStudy (xapproachesinfinity):

im not sure i have enough time

OpenStudy (anonymous):

then raise the power by +1 and solve it through again, righ?

OpenStudy (xapproachesinfinity):

No, you need a formula to do this using limits

OpenStudy (anonymous):

it's really quick!! you just have to check my answer :)

OpenStudy (anonymous):

what's the formula?

OpenStudy (xapproachesinfinity):

okay will see what we can do!

OpenStudy (xapproachesinfinity):

since this is geometric series there is a formula for Sn Sn=a1(1-r^n)/1-r Do you recognize this?

OpenStudy (anonymous):

ohhhh yeah

OpenStudy (anonymous):

also the next question is: What are the explicit equation and domain for a geometric sequence with a first term of 3 and a second term of -9? (5 points) an = 3(-12)n - 1; all integers where n ≥ 1 an = 3(-12)n - 1; all integers where n ≥ 0 an = 3(-3)n - 1; all integers where n ≥ 1 an = 3(-3)n - 1; all integers where n ≥ 0 i know it can't be the first two, so would the answer be D?

OpenStudy (xapproachesinfinity):

well that should be easy the first term is 3 the common ratio is -3 so the answer is c the first term should be 3 so all integers n greater than of equal 1

OpenStudy (xapproachesinfinity):

if it is d the meaning of common ratio will be last check for n=0

OpenStudy (xapproachesinfinity):

you will dealing some other ratio

OpenStudy (anonymous):

ok thanks i'll try

OpenStudy (xapproachesinfinity):

ok try to do the sum, i would like to see your result ^_^

OpenStudy (anonymous):

how exactly do i do the sum -_-'

OpenStudy (xapproachesinfinity):

use the formula that i wrote above. take the limit of that and see what it gives that should be your sum will you already know it has to be 1175

OpenStudy (xapproachesinfinity):

well*

OpenStudy (xapproachesinfinity):

is this algebra 2 course! i thought series are in precalc or calc lass?

OpenStudy (anonymous):

oh lol i was talking about the new problem -_-"

OpenStudy (anonymous):

it's alg 2

OpenStudy (xapproachesinfinity):

mmm.. that's weird! anyways, did you do the sum?

OpenStudy (anonymous):

yeah, it's 1175

OpenStudy (xapproachesinfinity):

okay! i think that answers your questions

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