Question

If this calculation continued forever, what would you expect the answer to be:

1/3 + 1/9 + 1/27 + 1/81 + 1/243 + ....

Answer 1

How can one do that?

Answer 2

Yeah

Answer 3

I must of forgotten how its done

Answer 4

I have one more question in relation to another math problem

Answer 5

Ok

Answer 6

Right

Answer 7

Ok

Answer 8

The power sign increases by one

Answer 9

The n

Answer 10

Does it have an answer because in the textbook the answer is 1/2

Answer 11

What does the R stand for and the 'a' and 'k'?

Answer 12

I sometimes skip class :)

Answer 13

Speaking from experience, @precal ? :D

Answer 14

I shouldn't that's why now I'm trying to focus more now on my studies

Answer 15

Two degrees in what subject?

Answer 16

Don't worry, @precal I'm a good (school) boy.
@Chad123 Don't you think it's odd that you can actually sum up an infinite number of addends and you still get an answer? :D

Answer 17

Maybe you did something else other than maths

Answer 18

Yeah, there should be k = 1 ad the bottom of the big sigma :D

Answer 19

if the exponent is n-1, it won't really matter :D

Answer 20

Okay, now I see it :D

Answer 21

I fix boredom by deriving :D
\[\huge S = \frac13 + \frac19 + \frac1{27}+\frac1{81}+\frac1{243}...\]

Answer 22

Here's an idea, @Chad123 Why don't we multiply both sides of the equation by 3?
\[\huge 3S = 1+\frac13+\frac19+\frac1{27}+\frac1{81}+\frac1{243}...\]

Answer 23

And then bring the 1 to the left side...
\[\huge 3S - 1 =\frac13 + \frac19 + \frac1{27}+\frac1{81}+\frac1{243}... \]

Answer 24

If you notice, the right-side is once again equal to S :D
\[\huge 3S -1 = S\]
Now it's just a linear equation in one variable. Get to it, champ, and solve for S :)

Answer 25

Because we started with this\[\huge S = \frac13 + \frac19 + \frac1{27}+\frac1{81}+\frac1{243}...\]

Answer 26

Eureka :)

Answer 27

Is there another way to the answer?

Answer 28

I agree to what you just said precal

Answer 29

Agreed :)
Allow me...
\[\large S=a+ar+ar^2+ar^3+ar^4...\]\[\large \frac{S}r=\frac{a}r+a+ar+ar^3+ar^4+ar^5...\]\[\large \frac{S}r-\frac{a}r=a+ar+ar^2+ar^3+ar^4...\]
And our usual result, the right side is once again equal to S :D
\[\large \frac1r(S-a)=S\]\[\large S-a=Sr\]\[\large S-Sr=a\]\[\large S(1-r)=a\]\[\huge \color{blue}{S=\frac{a}{1-r}}\]

Answer 30

So, plug in and be done with it,@Chad123 :D

Answer 31

a is the first term in the sequence(series) and r is the common ratio

Answer 32

a = 1/3 ?

Answer 33

Sounds logical to me now

Answer 34

What is the 3/4 doing there?

Answer 35

100 % sure

Answer 36

LOL
\[\huge \frac{a}{\frac23}=\frac12 \]\[\huge \frac23\times \frac{a}{\frac23}=\frac23\times\frac12 \]\[\huge \cancel{\frac23}\times \frac{a}{\cancel{\frac23}}=\frac{\cancel{2}}{3}\times\frac{1}{\cancel{2}} \]

Answer 37

Don't get dizzy, @precal XD

Answer 38

a is 1/3
?

Answer 39

Your mistake this time was multiplying 3/2 to both sides instead of 2/3 which is the correct factor to cancel out the denominator 2/3

Answer 40

That only applies when n starts at 0.

Answer 41

a is the first term... don't get bogged down into shortcuts :D