Question

What is the sum of the geometric sequence 2, 8, 32, … if there are 8 terms?
@kc_kennylau

Answer 1

1. Find its common ratio

Answer 2

2. Apply S=a(r^n-1)/(r-1)

Answer 3

thats 4 right?

Answer 4

yep :)

Answer 5

Then apply S=a(r^n-1)/(r-1) where a is the first term, r is the common ratio and n is the number of terms :D

Answer 6

10922.66

Answer 7

*cough* 43,690

Answer 8

:( how @kc_kennylau

Answer 9

sorry back

Answer 10

mind showing me what you got?

Answer 11

10922.66

Answer 12

I mean after you plugged it in, before calculating its value.

Answer 13

wait, I know what you did wrong

Answer 14

ok \[2(4^{8-1})/4-1\]

Answer 15

this is how i solved it

Answer 16

It's r^n-1 not r^(n-1) :)

Answer 17

PEMDAS

P-parentheses
E-exponents
MD-multiplication+division
AS-addition+subtraction

Answer 18

699050.66

Answer 19

so wait how does the equation look like

Answer 20

no, you must do it yourself, I cannot give you the answer

Answer 21

a(r^n-1)/(r-1)
\[\Large\dfrac{a(r^n-1)}{r-1}\]

Answer 22

Just plug in those darn values

Answer 23

sorry :(

Answer 24

not your fault :)

Answer 25

43690? finger crossed

Answer 26

finally......

Answer 27

*drumroll*

Answer 28

BINGO :DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

Answer 29

lol ;) thanks

Answer 30

no problem :)