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Precalculus
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Please help... picture attached
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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
there are a couple ways to do this, but i can only remember \[\sum_{k=1}^nk=\frac{n(n+1)}{2}\]
ooh i thought you had to actually add !
yes, it is the first one
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haha no.. thank you!! :)
i thought they wanted a nice closed form for this, which is not hard
yw
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