Question

Find a formula for Sn, (n is greater than or equal to 1)
4,8,16,32,64

@agent0smith

Answer 1

lol

Answer 2

Find the common ratio, divide one term by the term before it

Answer 3

i'm not saying anything until i find the answer to last nights question

Answer 4

^lol she posted it @satellite73 in the question.

Answer 5

btw this is a sequence, not a sum, maybe it is \(a_n\) rather than \(S_n\)

Answer 6

no its Sn.

Answer 7

Remember this from yesterday \[\Large S_n =a_1 \frac{ 1-r^n }{ 1-r }\]hopefully i remember that right.

Answer 8

ooh i see! i mean the one from last night!
how did you get it? did you read that business i sent, or did you figure it out some other way?

Answer 9

yes i remember that dang formula that i used all night lol

Answer 10

i figured it out another way.

Answer 11

:D well from 4,8,16,32,64 just find a1 and r.

Answer 12

A1 is 4 and r is 2?

Answer 13

at least there was a 60.21 in the answer
yes to your last question

Answer 14

yeah exactly...thanks again.

Answer 15

ho ho you are welcome, even though it wasn't quite right
you to this one now yes?

Answer 16

* you got this one

Answer 17

so im just using the same formula as last night?

Answer 18

yes exactly

Answer 19

if it was me, i would reverse the order top and bottom because \(r=2\) and \(2>1\) so i would write
\[4\frac{2^n-1}{2-1}=4(2^n-1)\]

Answer 20

@satellite73 I like how you mentioned to a pretty smart calculus student that 2>1 :D

Answer 21

lol thanks....but why would you switch them? and are you able to do it?

Answer 22

it is just if \(r>1\) you would have a negative number in the denominator (and also the numerator) since the denominator is \(1-r\)

Answer 23

and "pretty smart calc student" or no, it is always the case that
\[\frac{a-b}{c-d}=\frac{b-a}{d-c}\]

Answer 24

basically factoring out a negative \[\Large \frac{ a - b^n }{ a-b } = \frac{-( -a + b^n )}{ -( -a+b) }\]

Answer 25

so if \(r>1\) you know two things: the series does not converge, i.e. you cannot add up an infinite number
and also you probably want to use
\[a_1(\frac{r^n-1}{r-1})\]

Answer 26

especially if \(r=2\) because then \(r-1=1\)

Answer 27

@satellite73 i just don't think you found it funny like i did :P the 2>1 :D

Answer 28

no i liked it

Answer 29

okay lets drop it...obviously im not a genius in math but im trying my best...

Answer 30

here is a fact that you might want to commit to memory
\[1+2+4+8+16+..=\sum_{k=0}^n2^k=2^{n+1}-1\]
i think you are a math genius, i am just funnin
please do not take offence

Answer 31

so thats my answer? and sorry i dont really pick up on people being sarcastic when its over like facebook or text or open study...

Answer 32

\[\Large 4(2^n-1)\] this is the most simplified answer i think