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Mathematics 8 Online
OpenStudy (leozap1):

Will medal... Find the sum of the following series.

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.

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

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.

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

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.

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