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

\[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...\] so you know how to add this up ?

Answer 2

no i dont

Answer 3

\[a+ar+ar^2+ar^3+ar^4+...=\frac{a}{1-r}\] use
\[a=r=\frac{1}{3}\]

Answer 4

you can almost do it in your head
almost
what is one minus one third?

Answer 5

2/3

Answer 6

ok good
what is the reciprocal of \(\frac{2}{3}\) ?

Answer 7

3/2

Answer 8

and finally, what is \(\frac{3}{2}\times \frac{1}{3}\) ?

Answer 9

1/2

Answer 10

you win

Answer 11

im trying to understand the formula how it works

Answer 12

the formula i wrote?
\[\frac{a}{1-r}\]?

Answer 13

yes

Answer 14

you mean what to plug in for \(a\) and \(r\)?
or do you mean why the formula works?
or do you mean how you compute using the formula?

Answer 15

what is a and what is r?

Answer 16

\(a\) is the first term, and \(r\) is what you multiply the first term by to get the second term
what you multiply the second term by to get the third term, and so on

Answer 17

this formula only works if you have a sum that looks like
\[a+ar+ar^2+ar^3+ar^4+...\]

Answer 18

in your example you do
you have
\[\frac{1}{3}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \left(\frac{1}{3}\right)^2+...\]

Answer 19

in other words, you start with \(\frac{1}{3}\) as the first term and keep multiplying each successive term by \(\frac{1}{3}\)
it is not always the case that \(a=r\) but it is in your example

Answer 20

perfectly explained thanks alot @satellite73

Answer 21

yw

Answer 22

on e last question: how did you deduct the formula?

Answer 23

what if it is 1/3 + 1/9 - 1/27 + 1/81 - 1/243 + ...

Answer 24

@satellite73

Answer 25

hmm

Answer 26

drop off the first \(\frac{1}{3}\) and it looks like \(a=\frac{1}{9}\) and \(r=-\frac{1}{3}\)

Answer 27

compute
\[\frac{\frac{1}{9}}{1+\frac{1}{3}}\] and then add \(\frac{1}{3}\) to the result

Answer 28

could you please explain the formula?

Answer 29

why is a =1/9

Answer 30

1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + 1/64 - ...

Answer 31

this is the problem i cant get

Answer 32

you still there? i was off on a tangent

Answer 33

\[1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + 1/64... \] it alternates so \(r\) is negative

Answer 34

\(a=1,r=-\frac{1}{2}\)

Answer 35

compute as before
\[\frac{a}{1-r}\] with \(a=1,r=-\frac{1}{2}\) to get \(\frac{1}{1+\frac{1}{2}}\)

Answer 36

why is a=1 and r=-1/2?

Answer 37

but does the same logic apply for the first question @satellite73

Answer 38

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

Answer 39

will 1/3 be the first term (a=1/3)???

Answer 40

what will r be? 1st term multiply 2nd term (1/3 x 1/9)

Answer 41

@satellite73