Question

Help please . . .
(Picture Below)

Answer 1

@PeterPan

Answer 2

Can't see the number clearly; is that a 22?

Answer 3

z2

Answer 4

We can't approximate without knowing what z is...

Answer 5

Whats on the picture is my question. I don't even know why z is there. Thats why i posted it so i could get help.

Answer 6

We have to assume something, maybe that z isn't really there?

Answer 7

I guess. We can try it that way

Answer 8

\[\huge \sum_{x=0}^{15}2\left( \frac{1}{2} \right)^x\]

Answer 9

What kind of series is this?

Answer 10

Okay so what do i do ?

Answer 11

This is a geometric series, isn't it? We just keep multiplying 1/2

Answer 12

my answer choice is 4, 0 , 2 , 3 so it has to be a small number

Answer 13

2(1/2)=1 thats not an answer

Answer 14

Yes... so, use this formula
\[\huge S_n = a\left( \frac{1-r^n}{1-r} \right)\]
Still, you take a to be the first term in the series, and r is your common ratio, and n is the number of terms you add.

Go ahead and substitute, just like we did in your previous question.

Answer 15

2(1-whats r ? & n ?

Answer 16

n?
How many terms are we adding? We start from 0 and end at 15, so how many is that?

Answer 17

idk

Answer 18

breakk it down like last time so i can solve for my answer for some reason that helps me more

Answer 19

Well, you're gonna have to be able to do this by yourself, soon, so I'll just nudge you in the proper direction...
If you start at 0, then you're gonna have 16 counts to 15, right?
0,1,2,3,.......15

Answer 20

okay i see . continue

Answer 21

So your n = 16.
What's your common ratio?

Answer 22

so if n=16 do i put that in the formula

Answer 23

Yes, you do. Now, find r, your common ratio.
HINT:
A geometric progression always takes the form
\[\huge ar^x\]
In that case, your common ratio is r.

Answer 24

okay hold on

Answer 25

I'm back

Answer 26

@PeterPan

Answer 27

@terenzreignz