I really need help fast.. Use summation notation to express the sum. 2- 1/2 + 1/8 -..... + 1/2048
seems to get divided by 2 every term
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)
you tell me oops
and the sign alternates
what?
?
meanin +-+-
so mine is right?
The first term is 2. What are you dividing 2 by to get -1/2 ?
Oh you already have it, didn't see it. But your start at 2, not -8.
I have to go, can you just tell me if I am right please
Oh so is the side then: 2/4 ^n-1(-1)^n-1?
?????
one sec, I'm trying something out real quick
ok, then can u tell me if either of my choices are right?
isn't this a geometric series?
did u find out @jim_thompson5910
??????i need helppp
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)
Notice how when n = 7, a7 = 2*(-1/4)^(7-1) a7 = 1/2048
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??
would that be right?
This means that we stop at n = 7 So 2- 1/2 + 1/8 -..... + 1/2048 represented in summation notation is \[\Large \sum_{n=1}^{7}2 \left( -\frac{1}{4} \right )^{n-1} \]
well that would be a geometric progression with r =-2\[u _{n}=u_{1} \times r ^{n-1}\]\[u _{1}=2\rightarrow u _{n}=2\times (-2) ^{n-1}\]\[\rightarrow u _{n}=(-1)^{n-1}\times2^{n}\]
So @jim_thompson5910 that sigma you have on the bottom is correct? u sure?
I agree but are u sure?
yes I'm sure
You can plug in arbitrary values of n to generate any term you want
and you should get a term that's in the given sequence above
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