Question

evaluate the sum of the finite geometric series.
4+12+36+108+...+972

Answer 1

this is a geometric sequence which r = 3
972 = 4*3^(n-1)
n= 6
so sum = 4*(1-3^6)/1-3 = 4*(-728)/-2 = 1456

Answer 2

\[4(1+3+3^2+3^3+3^4+3^5)=4\times \frac{3^6-1}{3-1}\]

Answer 3

how does n=6

Answer 4

do you understand what i wrote?

Answer 5

no

Answer 6

i understand the equation better, but i need to know how to get N

Answer 7

which side, the left hand side of the equal sign or the right hand side? the left hand side i got by noting that each term had a common factor of 4, so i factored it out

Answer 8

that left me with
\[4(1+3+9+27+81+243)\] and these are all clearly powers of 3, so i rewrote it as powers of 3 as
\[4(1+3+3^2+3^3+3^4+3^5)\]

Answer 9

my teacher said to use this equation S = a*((1 - r^n)/(1-r))

Answer 10

then the right hand side i get by knowing that one more than 5 is 6, so it must be
\[4\times\frac{3^6-1}{3-1}\] and it is arithmetic from there on in

Answer 11

you teachers is telling you to use a formula, which is fine, that is what i did
but do not fret about what \(n\) or \( n +1\) is. \(n \) is the last power you see, in this case it was 5, so \(n + 1\) is \(5+1\) is 6

Answer 12

where you get the 5

Answer 13

look at the left hand side of the equal sign. i got it by computing

Answer 14

lets go slow
do you agree that we can factor out a 4 from each term and get
\[4(1+3+9+27+81+243)\]?

Answer 15

no

Answer 16

well if you cannot, then the answer is wrong

Answer 17

why you cant factor a 4 from 9

Answer 18

you started with this
\[4+12+36+108+...+972\] right?

Answer 19

oh yeah, lol my bad

Answer 20

so each term you see has a factor of 4
i factored it out of the whole thing. that is the first step

Answer 21

yeah, i know. I thought you meant 3,9, and 27.

Answer 22

oh no, i meant starting with what you wrote

Answer 23

yeah

Answer 24

so now again, we have
\[4(1+3+9+27+81+243)\] right?

Answer 25

yes

Answer 26

ok and each term after the 4 is a power of 3 right?

Answer 27

yes

Answer 28

so lets write it out again, this time as powers of 3
\[4(1+3+9+27+81+243)=4(1+3+3^2+3^3+3^4+3^5)\] is that part ok

Answer 29

yes

Answer 30

good. now you tell me, where does the five come from?

Answer 31

there is 5 terms

Answer 32

it is right there in the problem yes? it comes from the fact that when you write it in this form, the last term you are looking at is \(3^5\) i see it with my eyes.

Answer 33

yes

Answer 34

hope that answers the question "where does the five come from"
it is from computing what you have. then you can use the formula
\[4\times \frac{3^6-1}{3-1}=4\times \frac{729-1}{2}=2\times 728=1456\]

Answer 35

Yes, thanks

Answer 36

yw