Mathematics
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OpenStudy (anonymous):
Write the sum using summation notation, assuming the suggested pattern continues.
-4 + 5 + 14 + 23 + ... + 131
Please explain how you got your answer.
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OpenStudy (anonymous):
what are you adding each time?
OpenStudy (anonymous):
9
OpenStudy (anonymous):
right
so maybe you can use something like \[\sum_{k=0}^n -4+9k\]
OpenStudy (anonymous):
course you still got to figure out what \(n\) is
OpenStudy (anonymous):
Im assuming it's infinity considering the other choice I have is 15 and the 15th term would not be 131
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OpenStudy (anonymous):
\[-4+9n=131\] what is \(n\)?
OpenStudy (anonymous):
oh... 15....
OpenStudy (anonymous):
ahah. Thanks!
OpenStudy (anonymous):
ok so if \[\sum_{k=0}^{15} -4+9k\] is a choice, pick that one
OpenStudy (anonymous):
yw
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OpenStudy (anonymous):
It is. Mind helping me with another?
OpenStudy (anonymous):
go ahead and ask, maybe i can helpl
OpenStudy (anonymous):
Write the sum using summation notation, assuming the suggested pattern continues.
16 + 25 + 36 + 49 + ... + n^2 + ...
OpenStudy (anonymous):
does it really have \(...\)?
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
then just go with \[\sum_{n=4}^{\infty}n^2\]
OpenStudy (anonymous):
silly because you cannot add all this stuff up but whatever
OpenStudy (anonymous):
Why n=4 at the bottom?
OpenStudy (anonymous):
what number starts the summation ?
OpenStudy (anonymous):
Don't know. Would've said 16.
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OpenStudy (anonymous):
yes it is
OpenStudy (anonymous):
and what number squared gives 16?
OpenStudy (anonymous):
4. So It's the square root of the first term that goes on the bottom???
OpenStudy (anonymous):
\[\sum_{n=4}^{\infty}n^2=4^2+5^2+6^2+...\]
OpenStudy (anonymous):
each term is the square of some number
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OpenStudy (anonymous):
oh I see
OpenStudy (anonymous):
starting with 16, that is the square of 4
OpenStudy (anonymous):
Thanks again!
OpenStudy (anonymous):
yw again