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Mathematics 8 Online
OpenStudy (anonymous):

I really need help fast.. Use summation notation to express the sum. 2- 1/2 + 1/8 -..... + 1/2048

OpenStudy (anonymous):

seems to get divided by 2 every term

OpenStudy (anonymous):

Can I just tell you if my answer is right? I have 7 over sigma, n=1 below the sigma. to the side I have (-8 times (-0.25)^n)

OpenStudy (anonymous):

you tell me oops

OpenStudy (anonymous):

and the sign alternates

OpenStudy (anonymous):

what?

OpenStudy (anonymous):

?

OpenStudy (anonymous):

meanin +-+-

OpenStudy (anonymous):

so mine is right?

jimthompson5910 (jim_thompson5910):

The first term is 2. What are you dividing 2 by to get -1/2 ?

jimthompson5910 (jim_thompson5910):

Oh you already have it, didn't see it. But your start at 2, not -8.

OpenStudy (anonymous):

I have to go, can you just tell me if I am right please

OpenStudy (anonymous):

Oh so is the side then: 2/4 ^n-1(-1)^n-1?

OpenStudy (anonymous):

?????

jimthompson5910 (jim_thompson5910):

one sec, I'm trying something out real quick

OpenStudy (anonymous):

ok, then can u tell me if either of my choices are right?

OpenStudy (cwrw238):

isn't this a geometric series?

OpenStudy (anonymous):

did u find out @jim_thompson5910

OpenStudy (anonymous):

??????i need helppp

jimthompson5910 (jim_thompson5910):

Notice how you start at 2 Your next term is -1/2 The common ratio is then r = (term)/(previous term) = (-1/2)/2 = -1/4 So the geometric sequence is an = 2*(-1/4)^(n-1)

jimthompson5910 (jim_thompson5910):

Notice how when n = 7, a7 = 2*(-1/4)^(7-1) a7 = 1/2048

OpenStudy (anonymous):

So @jim_thompson5910 , it would be n=7 on top of sigma, n=1 on bottom. and the side is an=-2/4^n-1??

OpenStudy (anonymous):

would that be right?

jimthompson5910 (jim_thompson5910):

This means that we stop at n = 7 So 2- 1/2 + 1/8 -..... + 1/2048 represented in summation notation is \[\Large \sum_{n=1}^{7}2 \left( -\frac{1}{4} \right )^{n-1} \]

OpenStudy (anonymous):

well that would be a geometric progression with r =-2\[u _{n}=u_{1} \times r ^{n-1}\]\[u _{1}=2\rightarrow u _{n}=2\times (-2) ^{n-1}\]\[\rightarrow u _{n}=(-1)^{n-1}\times2^{n}\]

OpenStudy (anonymous):

So @jim_thompson5910 that sigma you have on the bottom is correct? u sure?

OpenStudy (anonymous):

I agree but are u sure?

jimthompson5910 (jim_thompson5910):

yes I'm sure

jimthompson5910 (jim_thompson5910):

You can plug in arbitrary values of n to generate any term you want

jimthompson5910 (jim_thompson5910):

and you should get a term that's in the given sequence above

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