Question

Express in Sigma Notation:
2+1+1/2+.....+1/128

Answer 1

I know it is a Geometric Series, and i know that to find the nth power is use the formula a_n=a_1(r)^n-1, but i dont know how to solve it after i plug in the values.

Answer 2

\[\sum_{i=1}^{9} 2 (1/2)^{n-1}\]

Answer 3

what is the limit you put above the sigma?

Answer 4

the reason i put the formula in my first comment is because i was trying to find a_n(the"place" of 1/128) is so that i know at what number place the series ends.

Answer 5

AKA the number above the sigma symbol

Answer 6

upper limit is 9

Answer 7

I wouldn't call it a limit

Answer 8

and did you use the formula\[a _{n}=a _{1}\times(r )^{n-1}\]

Answer 9

well not the limit, but the last number of that series

Answer 10

correct, the largest index

Answer 11

yesh. well could someone run me through the process of solving for "n" or the largest index? ......i dont know how to solve because "n" is in the exponent.....

Answer 12

um you are not solving for n, n is just a counter to let you know you will stop after n = 9

Answer 13

what i mean is how do i find our that n=9?

Answer 14

\(\sum_{i=0}^3 i = 0+1+2+3 = 6\)

Answer 15

\(\sum_{i=1}^{9} 2 (1/2)^{n-1}=2 (1/2)^0+2 (1/2)^1+2 (1/2)^2+2 (1/2)^3+......+2 (1/2)^9\)

Answer 16

so you just kept solving the series until you got to 1/128? see what if the largest index was 100? It would be impractical to keep solving until i reach the value at 100...so thats why i was using the Geometric Series sum formula to solve for the place or the highest index which in the formula is "n"

Answer 17

the question you posted was only asking to express in sigma notation not find the nth term

yes, if you had an index up to 100, it will be better to use the formula that solves for the nth term, but in this case you just asked to express it in sigma notation

Answer 18

but the thing is the last number is 1/128, but it only tells me the number, not the actually place which is apparently 9. Thats why i was using the formula to find "n" because that would give me the place of 1/128 which would end up being 9, and i need 9 because that is the highest index which i need to place above the sigma symbol. But i dont have the highest index...so how the heck do i get it?

Answer 19

To find n:
As you said, a_n = a_1 r^(n - 1).
Because a_n = 1/128, a_1 = 2 and r = 1/2,
1/128 = 2 * (1 / 2)^(n - 1)
1/256 = (1 / 2)^(n - 1) = 2^(1 - n)
Take log_2 on both sides. (If you don't have log_2, take log instead and then divide by log(2).)
-8 = 1 - n
n = 9

Answer 20

The key there is the log function. Whenever you need to solve for something in the exponent, try log.

Answer 21

is there any othe way?

Answer 22

other than the using the log to solve the equation?

Answer 23

There might be, but I am confident that log is the simplest.

One option would be to write a small script that does what you'd do manually, without log.

Answer 24

ok, for one last concern of mine. Is it correct to use the formula in the problem to find the highest index? Because everyone else didnt seem to care for it

Answer 25

Of course.

Answer 26

so it is correct to use the formula?

Answer 27

When you use the formula, you are clear about your reasoning and you get the right answer, so yes, it is correct.

Answer 28

lol ok thanks, sorry for the extra question, i failed the test and tommorrow is my last chance to redo it so i want to make sure im certain of everything haha

Answer 29

Good luck!

Answer 30

WAIT!'

Answer 31

in your solution you put that 1/128=2*(1/2)... Then after you multipled the 2 you rewrote as 1/256=...etc. How did you get 1/256?