Question

PLEASE HELP INFINITE GEOMETRIC SERIES
WILL GIVE MEDAL FAN

Answer 1

@Hero

Answer 2

@rebeccaxhawaii

Answer 3

Were you able to find the first four terms?

Answer 4

no

Answer 5

I don't know how

Answer 6

n=1 is your index, so that's where you start
what is the sum when plug in n=1 for n?

Answer 7

um

Answer 8

i'm not sure

Answer 9

what is (1/3) to power of 1?

Answer 10

1/3?

Answer 11

yeah, times -4 and then -1? Order of operations right? PEMDAS are you familiar?

Answer 12

yeah,

Answer 13

Where do you get -1?

Answer 14

@Directrix can you help me?

Answer 15

I got the terms 1, -1.333, -0.444444, -0.148

Answer 16

Are those wrong?

Answer 17

@phi @Directrix

Answer 18

Are you working on writing the first 4 terms?

Answer 19

yeah,

Answer 20

but it says n=1 so the first term is 1 right?

Answer 21

\[ \sum_{n=1}^{\infty} -4\left(\frac{ 1 }{ 3 }\right)^{n-1} \]

the first term is when n=1

Answer 22

ooh,

Answer 23

that means in the formula, you erase the "n" and put in 1 instead

Answer 24

so the first term is -4(?)

Answer 25

\[-4\left(\frac{ 1 }{ 3 }\right)^{1-1} \]

Answer 26

Ok, so then we get -4(1)

Answer 27

I assume you know 1-1 is 0
so it is
\[ -4\left(\frac{ 1 }{ 3 }\right)^0\]

Answer 28

ok,

Answer 29

anything to the "zero power" is 1, so the (⅓)^0 turns into 1
and -4*1 is -4

so -4 is the first term

Answer 30

could you check my other 3?

Answer 31

the next term will by -4 * (⅓)^1 which is -4 * ⅓ or -4/3

Answer 32

I got 2nd term: -1.3333333
3rd:-0.4444444
4th: -0.148

Answer 33

I would leave them fractions (unless they expect decimals .. which I doubt)

Answer 34

ooh

Answer 35

each term is ⅓ times the previous one
so you would have
-4 , -4/3 , -4/9 , -4/27

Answer 36

THANK YOU!!!

Answer 37

can you help me with b and c now?

Answer 38

I think that c is it has no sum since its infinite

Answer 39

the "bottom" is getting bigger and bigger. at some point for very large n
the bottom is huge and when you divide by a big number e.s. 1/1000000
the answer is a tiny number. in other words, the terms are "approaching zero"
and the series is converging

Answer 40

You and the Greeks thought that an infinite number of things diverged. But it turns out they can sum to less than a fixed value.

Answer 41

not all the time, but in this case we can find a value that the sum "approaches" in the limit

Answer 42

So the answer to b is that the converge(?)

Answer 43

yes

Answer 44

does it have a sum?

Answer 45

\[ \sum_{n=1}^\infty r^n = \frac{1}{1-r} \]
if r is less than 1

Answer 46

so what's the sum?

Answer 47

your problem is
\[ -4 \sum_{n=0}^{\infty} \left(\frac{ 1 }{ 3 }\right)^{n} \]

Answer 48

that's what I write for the sum in c?

Answer 49

no. the value of that infinite sum is 1/(1-r)
where r is ⅓ in this case

then multiply by -4 to get the final answer

Answer 50

sorry, I don't understand

Answer 51

There is a way to prove the series adds up to a number (in the limit)

the formula says the sum of
\[ \sum_{n=0}^{\infty}r^{n} = \frac{1}{1-r} \]

Answer 52

so the sum is -6?

Answer 53

I just did -4(1/1-1/3)

Answer 54

yes

the sum (⅓)^n from n=0 to infinity is 1/(1-⅓) = 1/2/3 = 3/2
and -4 * 3/2 = -6

Answer 55

yes

Answer 56

thank you so much

Answer 57

Do you think you could help me with another one?

Answer 58

btw, it might not have noticed, I changed the (n-1) exponent in your problem to n
and to compensate, I changed the starting n from n=1 to n=0
we get the same series, but now we can use the formula I posted , that goes from n=0 to infinity

Answer 59

@phi but are the answers still correct?

Answer 60

can you help me with another one?

Answer 61

yes, the answer is -6

Answer 62

could u help w/ antoher 1

Answer 63

you should make a new post