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OpenStudy (anonymous):
\[\sum_{n=1}^{\infty} (-1/2 )^{n-2}\]
OpenStudy (anonymous):
Okay, this is an infinite solution since the top is your upper limit and the bottom is the lower limit
OpenStudy (anonymous):
so,
-1/2^1-2 and so on - all the way to infinity - is going to be your answer
OpenStudy (anonymous):
\[\sum_{n=1}^{\infty} 3(0.4)^{n-1}\]
OpenStudy (anonymous):
that also and thanks keep explaining
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OpenStudy (anonymous):
1. -1/2^1-2 all the way to infinity (my answer for number 1?)
OpenStudy (anonymous):
Same thing for this equation. Replace the n with the lower limit (1) all the way to infinity. You don't have to show the work up to infinity, just lead up to it
OpenStudy (anonymous):
@Ket-kat so what would that be for number 2?
OpenStudy (anonymous):
?
OpenStudy (anonymous):
@Ket-kat whats the answer for 1 and 2, i'm confused.
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OpenStudy (anonymous):
hate that stuff
OpenStudy (anonymous):
@15Trev can you still help me though?
OpenStudy (anonymous):
Okay Like I said. In both equations on the left side of the stigma, replace the n with:
1,2,3,4,5,6,7,8,9,10.......... infinity
And solve the equations :)
OpenStudy (anonymous):
i think so
OpenStudy (anonymous):
Thanks
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OpenStudy (anonymous):
<3 No worries
OpenStudy (anonymous):
we did this last chapter
OpenStudy (anonymous):
FLVS? I'm doing it now xD
OpenStudy (anonymous):
Nice so its fresh for you
OpenStudy (anonymous):
k
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OpenStudy (anonymous):
so first u have to know that whats in parenths is ur rate