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Mathematics 12 Online
OpenStudy (anonymous):

Write the series in summation notation: -3 + 6 - 9 + 12 - 15

OpenStudy (kirbykirby):

A way to take care of the alternating signs (i.e. + - + - ) is to use the factor \(\large (-1)^n\) since: \(\large (-1)^1=-1\) \(\large (-1)^2=1\) \(\large (-1)^3 =-1\) \(\large (-1)^4=1\) etc.. Now you need to worry about going from 3 to 6 to 9 to 12 to 15. Hopefully you notice that this is simply multiplying by 3 every time. So, you can write this as \(3n\), were \(n\) will take on the values 1, 2, 3, 4, 5. So, the sum becomes: \[ \large \sum_{n=1}^5 (-1)^n(3n)\]

OpenStudy (anonymous):

how about ∑5k = 1[8(k + 1)]

OpenStudy (kirbykirby):

are you trying to evaluate tat?

OpenStudy (anonymous):

yes

OpenStudy (kirbykirby):

what is the summation running from? like going from k = ? to ?

OpenStudy (anonymous):

5??

OpenStudy (anonymous):

@MaimiGirl

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