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Find the sum of the following series.
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OpenStudy (leozap1):
|dw:1463436269305:dw|
OpenStudy (leozap1):
A. 240
B. 255
C. 210
D. 510
OpenStudy (leozap1):
It's n = 1 not n - 1. Sorry.
satellite73 (satellite73):
\[\huge \sum_{n=1}^{15}(2n+1)\]
OpenStudy (leozap1):
Yes, just like that.
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rebeccaxhawaii (rebeccaxhawaii):
Yes so you start at n=1 and end at n=15
rebeccaxhawaii (rebeccaxhawaii):
We have 15 terms actually.
OpenStudy (leozap1):
Okay, how do I plug it in?
rebeccaxhawaii (rebeccaxhawaii):
Getting ahead of myself - first we need to find n=1 and n=15
rebeccaxhawaii (rebeccaxhawaii):
To find this you will plug in 1 and 15 into 2n+1
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satellite73 (satellite73):
what did you get when you plug in 1 and then 15?
OpenStudy (leozap1):
I'm not sure how to plug it in. I've never done these problems before.
rebeccaxhawaii (rebeccaxhawaii):
when n=1 and n=15 plug it in. 2n+1 meaning plug 1 and 15 for n
OpenStudy (leozap1):
So 2(1)+15?
rebeccaxhawaii (rebeccaxhawaii):
No.
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rebeccaxhawaii (rebeccaxhawaii):
Plug in 1 and 15 for n. You will have 2 answers.
rebeccaxhawaii (rebeccaxhawaii):
2n+1 when n is 1 and 15
OpenStudy (leozap1):
Like this?
2(1) + 1 = 3
and
2(15) + 1 = 31
rebeccaxhawaii (rebeccaxhawaii):
There we go.
So a_1=3 and a_n=31
a_1 is the first terms and a_n represents the last term which in this case is a_15
n=15 because there is a total of 15 terms
rebeccaxhawaii (rebeccaxhawaii):
Plug all that into |dw:1463437146130:dw| the top/first formula
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OpenStudy (leozap1):
So 15/2 (3 + 31) ??
OpenStudy (leozap1):
That would give me 255.
rebeccaxhawaii (rebeccaxhawaii):
Great job! 255 is the sum of this arithmetic series.