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Precalculus 18 Online
OpenStudy (anonymous):

Please help... picture attached

OpenStudy (anonymous):

OpenStudy (anonymous):

8 + 14 + 20 + 26 + ... + 6n + 2 6n + 2 8 + 10 + 12 + 6n + 2 6n personally, i think it is the first, but i wanted to make sure

OpenStudy (anonymous):

there are a couple ways to do this, but i can only remember \[\sum_{k=1}^nk=\frac{n(n+1)}{2}\]

OpenStudy (anonymous):

ooh i thought you had to actually add !

OpenStudy (anonymous):

yes, it is the first one

OpenStudy (anonymous):

haha no.. thank you!! :)

OpenStudy (anonymous):

i thought they wanted a nice closed form for this, which is not hard

OpenStudy (anonymous):

yw

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