Question

use a graphing utility to find the sum of each geometric sequence.

(1/4) + (2/4) + (2^2/4) + (2^3/4) + ... + (2^14/4)

Answer 1

Please tell me you didn't try to do this manually :D

Answer 2

lol im very close to....

Answer 3

It's a geometric series, and if written in the convenient sigma notation, it'd look something like this...
\[\huge \sum_{n=1}^{15}\frac{2^{n-1}}4\]

Answer 4

now the answer is in the back of the book and it shows me everything to put into my calculator and gives me the answer but it doesn't tell me where the buttons are im supposed to be pressing

Answer 5

I wrote that formula down next to the problem

Answer 6

Oh... can't help you with that, I have no idea what calculator you're using...
There is a formula for geometric series...though...

Answer 7

the issue im having is it says to solve by using a calculator and I don't know how to put it into that

Answer 8

im using a TI84 plus

Answer 9

Yikes... better inquire as to who has such a calculator, then :D
Good luck :)

Answer 10

8191.75

Answer 11

the answer is 8191.75 I have the answer and what is put into the calculator to get it I just don't know what to put down in the calculator

Answer 12

yes!

Answer 13

now what do you put into your calculator I have a TI 84

Answer 14

the sum of GP series is \[a(r ^{n}-1)/(r-1)\]

here a is 2, r=2, n=15...
then divide it by 8 as the equation is
\[\sum_{n=1}^{15} 2^{n}/8\]

Answer 15

whoa there... a is not 2...
a is the first term in this progression...

Answer 16

i dont have a have a graph calculator...
@terenzreignz , in this equation, 'a' a.k.a 'the first term' is 2

Answer 17

No... it's 1/4

Answer 18

okay wait why is the number of the summation sign 15 and not 14

Answer 19

Because there would be 15 terms, @d2thej

Answer 20

if the number stops at 14 wouldn't there only be 14?

Answer 21

do you just add one to that number just for the heck of it?

Answer 22

simplifiy the equation you have written in the previous answer.......2^(n-1)/4......make it (2^n)/8

Answer 23

It does stop at 14 (the exponent of 2)
however, it STARTED at 0.
See, the first term is 1/4
OR
\[\huge \frac{2^0}{4}\]
So there are, in fact, 15 terms...

Answer 24

hmm I guess ill jus take your word for it, im not seeing it

Answer 25

@jiraya what do you mean what am I simplifying

Answer 26

Okay... visualize :D
Let's stop it at 3, for instance...
\[\huge \frac{1}{4}\quad\frac{2}{4}\quad\frac{2^2}{4}\quad\frac{2^3}{4}\]

See, when we stop at 3, we, in fact, have 4 terms...

similar logic, basically

Answer 27

ahhhhhh okay

Answer 28

so its like you count how many are raised to a power and the one that isn't I guess?

Answer 29

Of course :)
We're counting TERMS and that's that, no matter what they look like.

But if you really must have uniformity...
\[\huge \frac{2^0}{4}\quad\frac{2^1}{4}\quad\frac{2^2}{4}\quad\frac{2^3}{4}...\quad\frac{2^{14}}{4}\]

And now it's clear to you that there are 15 terms?
Because we stopped at 14, but we STARTED at 0 (not 1)

Answer 30

yes! I get it now lol thanks, so what buttons am I pushing exactly to put it in my calculator lol

Answer 31

That... you're gonna have to find another way to do that, I don't have your calculator, sorry bro :D

Answer 32

nvm I seen what @jiraya did

Answer 33

lol thanks though

Answer 34

okay wait now,

Answer 35

how would I know to put 8 on the bottom instead of 4

Answer 36

I guess it's just to make the summation look nicer... hang on...

Answer 37

okay I got more questions if you don't mind lol

Answer 38

I have to go in a bit... maybe just one :)

Answer 39

okay same instructions except they give me this: \[\sum_{n=1}^{15} (2/3)^n\]

Answer 40

Even better (that is, if you're trying to find the sum)
You can just plug in directly :D

Answer 41

Obviously your common ratio r = 2/3
and your a (first term) would be (2/3)^1 = 2/3
and just plug in

Answer 42

it doesn't come out to what the answer is in back of the book lol

Answer 43

What formula did you use?

Answer 44

the summation one

Answer 45

ohhh wait nvm I got it

Answer 46

This formula...
\[\Large \frac{a(1-r^n)}{1-r}\]

Answer 47

it depended on how I put it into the calculator

Answer 48

i don't know if you've ever used one of these before but the 2/3 had to be in its separate parenthesis other than what the equation gives you

Answer 49

thanks for your help if you need to go that's cool lol can i add you as a friend on here?

Answer 50

Seems you're way ahead of me :)
I really need to go now, though, try to find someone who knows your calculator inside and out :D
I'll be seeing you :)
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Terence out