Question

Geometric sum. Should be a fun one :).
Evaluate the sum 1/2 + 1/6 + 1/18 + 1/54 + ... + 1/1458

Answer 1

I know how to put other problems like this into an equation to get the sum. The difference with this one is it's not telling me how many terms there are in between.

Answer 2

I do realize somewhere along the lines it's multiplying each term by 1/3

Answer 3

Geometric sum equation \[a1(\frac{1-r^n }{1-r})\]

Answer 4

I agree with that :)

Answer 5

Problem is need to find n I think.

Answer 6

a1=1st term
r=your common ration (1/3 like you said)
n=number of terms

Answer 7

it contain total 7 terams since (3^6)*2= 1458
so find sum using the eqn

Answer 8

Well thats where we go to the geometric sequence formula: \[a _{n}=a_{1}r^{n-1}\]

Answer 9

\[a _{n}\]=the last term in the sequence and we already know that to be 1458 and all the other variables are the same like in the last formula

Answer 10

the answer is 1093/1458

Answer 11

So I see how you plugged it in. We need to solve for that n right?

Answer 12

So you end up with log2(1/729) = n-1 ?

Answer 13

No, sorry my bad

Answer 14

Hold up I put the worng number up \[\frac{ 1 }{ 1458 }=\frac{ 1 }{ 2 }*\frac{ 1 }{ 3 }^{(n-1)}\]

Answer 15

It's ok :), take your time. I appreciate the help.

Answer 16

\[\frac{ 1 }{ 729 }=\frac{ 1 }{ 3 }^{n-1}\]

Answer 17

This makes makes more sense to me now, lol

Answer 18

\[\log_{\frac{1}{3}} \frac{ 1 }{ 729 }=\log_{\frac{1}{3}}\frac{1}{3}^{n-1}\]

Answer 19

\[6=n-1\]
\[7=n\]

Answer 20

OK we got our "n" now we can proceed :)

Answer 21

very cool :).

Answer 22

@Jhannybean Its called the change of base formula

Answer 23

Ok, lol that ws easy anywyas back to @bigmatt
\[\frac{1}{2}(\frac{1-1/3^{7}}{1-(1/3)})\]

Answer 24

We get 1093/1458 I think.

Answer 25

That's it, lol I good, I didn;t want to compute that, unless you don't see it, anything else?

Answer 26

Ya, that's it! I just needed help with finding n and you did great :) Thank you!

Answer 27

Np, anytime