Question

sum_{i=1}^{B}C
Rewrite the sume 1/2-1/4+1/8-1/16+1/32-1/64+1/128.
B is 7. What is C?
Please help!

Answer 1

im a little confused by the notation... is the the question?\[\sum_{i=1}^{B}C=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots\]and you need to figure out what C is?

Answer 2

yes that is exactly it!

Answer 3

Do see any type of pattern in the denominators?\[2,4,8,16,\ldots \]

Answer 4

yes I see that it is doubled

Answer 5

Right! so we could rewrite them as:\[2,4,8,16,32,\ldots\rightarrow 2^1,2^2,2^3,2^4,2^5\ldots\]

Answer 6

Oh okay! So what would i be exactly? 2^i?

Answer 7

sorry i mean C

Answer 8

if the sequence was what I posted above, you would be correct, but the sequence of your actual problem is:\[\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\ldots\]

Answer 9

so we still need a way to tackle the plus minus plus minus...

Answer 10

Oh okay so maybe -1/i^2 ?

Answer 11

what you had before was closer. Since:\[2^i\Rightarrow 2^1,2^2,2^3,\ldots\]and we want those numbers in the denominator, we should put:\[\frac{1}{2^i}=\left(\frac{1}{2}\right)^i\Rightarrow \frac{1}{2},\frac{1}{2^2},\frac{1}{2^3}\ldots\]

Answer 12

Oh okay! So it would simply be (1/2)^i?

Answer 13

I'm a little confused since it has the negative..

Answer 14

yeah...we still have to deal with the negative... lets look at this sequence:\[1,-1,1,-1,1,-1\ldots\]mabye we can come up with a formula for it.

Answer 15

this is where I'm stuck :(

Answer 16

hmm...what would I multiply 1 by to get -1?

what would I multiply -1 by to get 1?

Answer 17

-1

Answer 18

would it be -(1/2)^i ?

Answer 19

so close!! it would be:\[\left(-\frac{1}{2}\right)^i\]

Answer 20

Oh that totally makes sense now!! THank you SO much for your help! :)