Question

Find the sum of (-3)^(n-1) divided by 4^n. This is a geom. series. When I simplify it, im confused on what my "a_1" should be.

Answer 1

\[\frac{(-3)^{n-1}}{4^{n}} = \]

Answer 2

all I can see is that, by simplifying, you can get (-1/3) * (-3/4)^n. Does that help in any way?

Answer 3

I have that too!!! Im going over an old test, and i got that answer wrong. The teacher said that a_1 should "1/4". But now that Im looking at it, i think it should be "1/3".

Answer 4

1/4? Hmm... it could possibly be 4 as well I guess... 4 * (-3/4)^(n-1) but 1/4 doesn't look right

Answer 5

I'm no expert tho... Maybe someone else could offer a second opinion?

Answer 6

Yeaah, im wavering between the two. I agreed with 1/4 at first. But it looks weird now.

Answer 7

The sum w/ (1/3) gives me 1.333 and the sum w/ (1/4) just gies me 1. Lol, they both kinda converge

Answer 8

Series are the part of the semester which I've slept through... It's amazing I had even passed that course in the first place, so how do you determine when a serie converges? (if it's not too long to explain)

Answer 9

My test on this stuff is today, so i HOPE i can even explain!
If |r|<1, then it converges to Sum= a_1/1-r.
If |r| >= 1 then the above geometric series diverges.

Answer 10

ohhhhh, hold on, 1/4 is good too.... dunno why I screwed up that one, but you can get (-1/3)(-3/4)^n or (1/4)(-3/4)^(n-1)... So 4 * (-3/4)^(n-1) is WRONG and I do apologize. On your bottom term, 4^n = 4(4^(n-1))... I dunno why you would pick to have your exponent as n-1 instead of n, but I hope you do ;-)

Answer 11

Yes, you are pulling out the "4", which means you are subtracting. thats why you have the "n-1". Doing so makes the exponents equal. so (-3^n-1)/(4^n-1). You see? lol

Answer 12

yeah, I guess, but I would naturally prefer pulling out the -3 from the top bit, to make the exponent n. It just looks cleaner with n, but it's probably not how to do solve a geometric series... anyways, I think I've caused more confusion than anything here (at least, in my brain I did)

Answer 13

THIS is why I dont like math. Thanks anyways. Since you had to take this course, what are you majoring in??

Answer 14

oh, it's been at least 10 years. I live in Quebec and this course was part of what we call a "pre-university" degree (it's a weird education system here), taken in college. I then left math behind to try my hand at biology, which was even more boring ;-)

Answer 15

Some schools here in Georgia & US have that. Im doing electrical engineering and im bored as well. Thanks for your help!!!!!!!!

Answer 16

I'm not sure I'm doing this correctly, but I seem to get that this serie converges to 1/7... I'm doing this :
s = 1/4 - 3/16 +9/64 - 27/256 + ... + 1/4(-3/4)^(n-1)
-3s/4 = -3/16 + 9/64 - 27/256 + ... + 1/4(-3/4)^n

s-(-3/s4) = 1/4 + 1/4(-3/4)^n
7s/4 = 1/4(1 + (-3/4)^n)
s = 1/7 ( 1 +(-3/4)^n)
if n tends to infinity, s = 1/7, right?

Answer 17

oh, when you said you had gotten 1 with 1/4, I think you might've done s - 3/4s, correct?
s/4 = 1/4(1+(-3/4)^n)
s = 1+(-3/4)^n
s = 1

Answer 18

Its righte, but You definintely did it the hard way. S= (1/4) / (1-(3/4))

Use formula: 1st term / (1- |r|)

Answer 19

are you sure that you need the absolute value of r in your formula? because Wolfram seems to tell that this series converges to 1/7 as well, and using absolute value doesn't make sense mathematically :-S

Answer 20

Hey, thats what the book and the teacher says. I have learned no other way. Because we have r= (-3/4), So |r|=|-3/4|= 3/4 <1, it converges. Then use the sum thingy

Answer 21

oh, yes, to determine that it converges, you definitely need to use the absolute value, but to determine the actual value towards which it converges, I think that you don't need it (at least, I couldn't find it in google, but I did find instance of s = a_1/(1-r) tho)

Answer 22

and that's why I hate formulas, because if you remember them wrong or if you learnt them wrong, you can be wrong... @experimentX @foolformath guys (or girls, I dunno), can you help us out here please, or summon people which you know could help?

Answer 23

Yes, I said that. I never said that you need |r| in the sum thingy. You only need |r| to see if it converges.

Answer 24

(-3)^(n-1) / 4^n = (-1)^(n-1)*(3/4)^n/3 <--- looks easier now.

Answer 25

both are definitely guys @m_charron2

Answer 26

\[ a_1=\frac 1 4 \]

Answer 27

\( \frac{(-1)^{n-1}}{3} (\frac{3}{4})^n\)
put the values of n, you should be easily to find.

Answer 28

a1 = 1/4
a2 = -3/16
a3 = +9/64
..
..

Answer 29

I will like to represent the n-th term as:\[a_n = \frac 14 \times \left(-\frac 34\right)^{n-1} \]

Answer 30

now \(|r| = \frac 3 4 \lt 1 \), hence the series converges.

Answer 31

Thanks everyone!