Question

8. Find S8 for the geometric series 3 + -6 + 12 + -24 +…
a. -255
b. 768
c. -96
d. 192

Answer 1

easy use

ar0+ar1+ar2

so a is 3

Answer 2

r is
-6=3*r r=-2

Answer 3

so then 3*(-2)^57

Answer 4

no wait i thought you meant term lol

Answer 5

but stil a and r are valid

Answer 6

\[\sum_{n=0}^73\times (-2)^n=3\times \frac{(-2)^8-1}{-2-1}\]

Answer 7

use this formula

\[{\frac {a{r}^{n+1}}{r-1}}-{\frac {a}{r-1}} \]
in it i showed you a and r

n is the ith_term-1 which is 58-1 in your case

Answer 8

what timo86m said is correct, \(a=3,r=-2\) formula to use is
\[\sum_{k=0}^nar^k=a\frac{r^{n+1}-1}{r-1}\]

Answer 9

i get \(-255\)

Answer 10

\[\sum_{n=0}^73\times (-2)^n=3\times \frac{(-2)^8-1}{-2-1}=3\frac{256-1}{-3}=-255\]