Question

find the sum of the finite geometric series.... 4+ 4/3+ 4/9+ 4/27+...+ 4/6561 i don't know if i'm using the correct formula?

Answer 1

I'm using: sn=\[\frac{ 1-r^n }{ 1-r }\]

Answer 2

then times a1, with a1 is the first term

Answer 3

@morningperson1991
Is your doubt cleared? :)

Answer 4

oops! i forgot to add that to the formula i put up. it still looks like an unbelievably small number. my r=1/3 and my a1=4. i tried to find n by using the formula: a1*\[r^n-1\] but i'm stuck in that step. suggestions?

Answer 5

I guess the trouble is with finding n..my answer is extremely small

Answer 6

Use
ar^n-1 = 4/6561
a = 4
r = 1/3
Find n like that. :)

Answer 7

i believe thats the same formula i've been using @AkashdeepDeb :/ i'll try again now

Answer 8

Hahah...Yeah :D

Answer 9

since i'm multiplying 4 by the r term. should i divide 4 by each side to get rid of it? my calculator is giving me crazy numbers that don't turn into fractions...

Answer 10

Hold on let me try it.

Answer 11

@AkashdeepDeb are you getting the (1/3)^n-1 = 1.5241E-4 ? this number is crazy.. i can't turn it into a fraction

Answer 12

hold on hold on :P

Answer 13

@akashdeepDeb do you know of another formula i can use to find n?

Answer 14

what formula are you using to find n ?

Answer 15

I'm using \[a _{1}r^n-1=\frac{ 4 }{ 6561 }\]

Answer 16

you mean
\[ a _{1}r^{n-1}=\frac{ 4 }{ 6561 } \]

Answer 17

oh yes! i've been using that formula. that was a typo while i input the formula. sorry. I'm just stuck on the n part... i get crazy small numbers that i cant turn into fractions. any suggestions? @phi

Answer 18

with a1= 4 you get
\[ 4r^{n-1}=\frac{ 4 }{ 6561 } \\ \left(\frac{1}{3}\right)^{n-1}=\frac{ 1 }{ 6561 }\]

Answer 19

we can flip both sides, just so it looks nicer
\[ 3^{n-1}= 6561\]
How would you solve this? I would use logs:
\[ (n-1) \log(3) = \log(6561) \\ (n-1) = \frac{\log(6561)}{\log(3)} \\ n = \frac{\log(6561)}{\log(3)}+1\]

Answer 20

@phi that looks very understandable except for the first step.. when you divided both sides by 4. how did it not change the denominator? it only divided by the numerator. can you please explain? sorry, im trying to understand this problem :/ Thank you so much for those steps!

Answer 21

are you asking about
\[ 4r^{n-1}=\frac{ 4 }{ 6561 } \] ?
multiply both sides by 1/4 (or divide both sides by 4)
you get
\[ r^{n-1}=\frac{ \cancel{4} }{ 6561 }\cdot \frac{1}{\cancel{4}} = \frac{ 1 }{ 6561 } \]

Answer 22

Sorry I was out.

Answer 23

ohhh! that's embarrassing... @phi thank you so much!

Just to verify. can your calculator turn the final sn answer into fractions? because my couldn't. Once i put in n=9 into the geometric summation formula i got 5.99969. I don't know if my teacher will allow me to round that up to 6. but just to make sure, did you get the same answer?
@AkashdeepDeb , that's alright :) thank you for trying to help!

Answer 24

r = 1/3
So
1/3^n . 3 = 1/6561
1/3^n = 1/19863
n = 9
Now find the sum.

Answer 25

you can use wolfram
http://www.wolframalpha.com/input/?i=sum+4*%281%2F3%29%5Ei++for+i%3D0+to+8

it found 39364/6561

Answer 26

my calculator can change the decimal into that ugly fraction

Answer 27

@phi thank you so much! :) My calculator wouldn't change anything in this problem into fractions.. its a TI=84.. but thanks anyway :)

Answer 28

@AkashdeepDeb thank you too! :)

Answer 29

:)