Mathematics
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OpenStudy (anonymous):
Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15.
n an
1 5
2 −10
3 20
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OpenStudy (anonymous):
@math&ing001
OpenStudy (math&ing001):
\[\sum_{n=4}^{15}(-1)^{n-1}2a _{n-1}\] with a1=5
OpenStudy (math&ing001):
Is that what you're looking for ?
OpenStudy (anonymous):
\[\sum_{n=4}^{15}5(-2)^{n+1}\]
OpenStudy (anonymous):
this is the option most close to yours
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OpenStudy (math&ing001):
Let me check
OpenStudy (anonymous):
ok til then i'll write the other options for you
OpenStudy (math&ing001):
\[\sum_{n=4}^{15}5(-2)^{n-1}\] Does this figure in your choices ?
OpenStudy (anonymous):
\[\sum_{n=1}^{15}5(-2)^{n-2} \]
\[\sum_{n=1}^{15}5(-2n)\]
\[\sum_{n=1}^{15}5(-2n)^{n+1}\]
OpenStudy (anonymous):
these are the rest of the options
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OpenStudy (math&ing001):
Look at my answer above.
OpenStudy (math&ing001):
Cause from the sequence given it should be n-1 not n+1
OpenStudy (anonymous):
so its the one i typed firrst
OpenStudy (math&ing001):
Yeah, except 2 should be at the power of n-1
OpenStudy (math&ing001):
I mean (-2)^(n-1)
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OpenStudy (anonymous):
ok so its the one n=4? sorry I know I'm annoying
OpenStudy (math&ing001):
yes :)
OpenStudy (anonymous):
Thank You!!
OpenStudy (math&ing001):
Welcome !
OpenStudy (anonymous):
FLVS students:
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OpenStudy (math&ing001):
Actually, Malice it's the first choice : \(\sum_{n=4}^{n=15} 5(-2)^{n-1} \)