Question

sum of the infinite geometric series (1/2)+(1/4)+(1/8)+(1/16)+...

Answer 1

\[a_1+a_2+a_3+a_4+...\]
\[r=\frac{n_2}{n_1}\]
\[S_\infty=\frac{a_1}{1-r}\]

Answer 2

how do you find r though?

Answer 3

Use 1/4 divide 1/2 gets you r

Answer 4

picture above shows the answer without any computation
just thought i would mention it

Answer 5

Lol how do you use that diagram to answer

Answer 6

fill it in
you can see that it will fill up the whole unit square, the further you go
so the answer is one

Answer 7

also you might note that in binary
\[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=0.11111...\]

Answer 8

ohhhh i see thank you both :D

Answer 9

and just like if you have base ten \(0.999...=1\) so in base two \(0.11111...=1\)

Answer 10

ah

Answer 11

but really you are supposed to do what @.Sam. said
\[\frac{\frac{1}{2}}{1-\frac{1}{2}}\]

Answer 12

you might also note that in base 3
\[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...=0.1111111\]

Answer 13

that's interesting :D thank you!!

Answer 14

yw

Answer 15

I like your different approach