Question

Can't find part B!
For the sequence An = 8/3^n,
A) its 5th partial sum S_5=968/243
B)its nth partial sum Sn = ?

Answer 1

I'm assuming the sequence is

\[\large A_{n} = \frac{8}{3^n}\]

Answer 2

That is correct.

Answer 3

Break it up and rewrite it to get

\[\large A_{n} = \frac{8*1}{3^n}\]

\[\large A_{n} = \frac{8*1}{1*3^n}\]

\[\large A_{n} = \frac{8}{1}*\frac{1}{3^n}\]

\[\large A_{n} = 8*\frac{1}{3^n}\]

\[\large A_{n} = 8*\frac{1^n}{3^n}\]

\[\large A_{n} = 8*\left(\frac{1}{3}\right)^n\]

Answer 4

What do you notice about the last equation?

Answer 5

I'm not sure, it just seems reminiscent of a the formula used to find Sn...or one of the others.

Answer 6

actually, let me rewrite it in a slightly different way

one sec

Answer 7

\[\large A_{n} = \frac{8*1}{3^n}\]

\[\large A_{n} = \frac{8*1}{1*3^n}\]

\[\large A_{n} = \frac{8}{1}*\frac{1}{3^n}\]

\[\large A_{n} = 8*\frac{1}{3^n}\]

\[\large A_{n} = 8*\frac{1^n}{3^n}\]

\[\large A_{n} = 8*\left(\frac{1}{3}\right)^n\]

\[\large A_{n} = 8*\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{n-1}\]

\[\large A_{n} = \frac{8}{3}\left(\frac{1}{3}\right)^{n-1}\]

Answer 8

What form is that last equation in?

Answer 9

I'm sorry, I'm really missing this...

Answer 10

what sequences have you studied so far?

Answer 11

My teacher has just started on arithmetic series and sequences and he mentioned geometric ones.

Answer 12

what form is an arithmetic sequence in

Answer 13

\[a_n=a_1+(n-1)d?\]

Answer 14

good, how about geometric

Answer 15

\[a_k+1=a_kr\]

Answer 16

With the one subbed a long with the k.

Answer 17

do you know of any other way to express a geometric sequence in general?

Answer 18

I'm not sure if it pertains, but I had read in my book of : xn = ar(n-1)

Answer 19

r^(n-1)*

Answer 20

good

Answer 21

look at the last equation I wrote above

notice how it matches up with that form you just wrote

Answer 22

Yes, it certainly does.

Answer 23

so what are 'a' and 'r'

Answer 24

a = 8/3 and r = (1/3)^(n-1)

Answer 25

close, a = 8/3 is correct

Answer 26

but r = (1/3)^(n-1) is NOT correct

Answer 27

Is it because r stands for common ratio? Because if so, it'd be (1/2)^(n-1)

Answer 28

it should be r = 1/3

do you see how/why?

Answer 29

Yeah, I think I understand why.

Answer 30

now that you know 'a' and 'r', how can you use them to find the nth partial sum?

Answer 31

Is there actually a formula for that?

Answer 32

yes, one sec

Answer 33

ok

Answer 34

check out this page

http://www.regentsprep.org/regents/math/algtrig/ATP2/GeoSeq.htm

Answer 35

I see the Sn formula, if I were to plug in my values, it should look something like this:

Answer 36

\[(8/3 (1-1/3)^n)/(1-1/3)\]

Answer 37

Which is 2^(n+2)3^-n

Answer 38

one sec

Answer 39

ok this is what I'm getting


\[\large S_{n} = \frac{a(1-r^n)}{1-r}\]


\[\large S_{n} = a\frac{1-r^n}{1-r}\]


\[\large S_{n} = \frac{8}{3}\frac{1-(1/3)^n}{1-1/3}\]


\[\large S_{n} = \frac{8}{3}\frac{1-(1/3)^n}{2/3}\]


\[\large S_{n} = \frac{8}{3}(1-(1/3)^n)*\frac{3}{2}\]


\[\large S_{n} = \frac{8}{3}*\frac{3}{2}(1-(1/3)^n)\]


\[\large S_{n} = \frac{8*3}{3*2}(1-(1/3)^n)\]


\[\large S_{n} = \frac{24}{6}(1-(1/3)^n)\]


\[\large S_{n} = 4(1-(1/3)^n)\]

Answer 40

Why did you move the a out to the front though?