Question

7. Find S8 for the geometric series 256 + 64 + 16 + 4 +…
a. 357.62
b. 341. 33
c. 338. 25
d. 324.13

Answer 1

theres a nice little formula, but you need to know the common ratio to use it

Answer 2

if i recall it cirrectly, it goes\[\frac{1-r^n}{1-r}\]

Answer 3

the brute math method is to list all 8 terms and add them up, but youd stil need to determine the common ratio

Answer 4

r = 1/4

Answer 5

use that and find

Answer 6

how does r = 1/4 ?

Answer 7

WHAT AMIRSTRE SAID I MEAN WHYYYYYYYYY WOULD YOU DO THAT YAHOO WHY

Answer 8

since it is in gp

Answer 9

r = 64/256

Answer 10

=1/4

Answer 11

@amistre64

Answer 12

in general, each new term is the product of the one before it and some common value (r)

\[a_{n+1}=a_n*r\]

to find r, we divide both dies by an

\[\frac{a_{n+1}}{a_n}=r\]and that determines the common ratio

Answer 13

in gp each term is multiplied........

Answer 14

take any 2 of the numbers that are in a row, 16 and 4 and stack them last over first:

4/16 = r = 1/4

Answer 15

to test this r out, lets dbl chk:

256*1/4 = 64
64*1/4 = 16
16*1/4 = 4
4*1/4 = 1
1*1/4 = 1/4
1/4 * 1/4 = 1/16
etc ....

Answer 16

the rest is to plug everything into the formula for summation:\[\frac{1-r^n}{1-r}\to\
\frac{1-(1/4)^8}{1-1/4}=\frac{something}{3/4}\]

Answer 17

It is a decimal though which is not one of the answers....

Answer 18

fractions are decimals, you just have to simplify it

Answer 19

a calculator helps at times

Answer 20

1/4 = .25 might be useful

Answer 21

we might also want to make it easier, add up the inter parts and just use summation onthe fractions remaining

Answer 22

i think there is a tiny mistake here

Answer 23

256*1/4 = 64
64*1/4 = 16
16*1/4 = 4
4*1/4 = 1
.................................. 85

n=4 left

Answer 24

where??

Answer 25

i forgot a 256 along the way eh

Answer 26

yeah there

Answer 27

256
85
----
341 + n=3 left

Answer 28

341+ something smaller than 1 would seem to be b right?

Answer 29

I'm taking that! thank you for extensively helping me lol

Answer 30

lol, youre welcome ;) good luck

Answer 31

yes it is definitely 341 and a bit

Answer 32

\[\sum_{k=0}^7256(\frac{1}{4})^n=256\times \frac{1-(\frac{1}{4})^8}{\frac{3}{4}}\]