Find the sum of the following series. (shown below) A. 240 B. 255 C. 210 D. 510

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Find the sum of the following series. (shown below) A. 240 B. 255 C. 210 D. 510

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1434376110688:dw|
@aloud that's not one of my answers @phi PLEASE HELP!
no you ditz it's not one of them don't you get it?! it's not one of the answer choices!

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Other answers:

  • phi
rewrite the problem \[\sum_{1}^{15}(2n+1)= 2\sum_{1}^{15}n+\sum_{1}^{15}1\]
  • phi
There is a formula to add up the numbers from 1 to n do you know it? and the second sum means add up 15 1's (which hopefully you know how to do)
Mmm.... what would you say if I said I knew it but I couldn't remember it? :p
  • phi
Gauss (famous mathematician), when a young kid in school was given the problem of adding up the numbers from 1 to 100, and he (clever fellow) saw a way to do it quickly. \[\sum_{k=1}^{n}k= \frac{ n(n+1) }{ 2 }\]
  • phi
your problem has n=15 so use that formula with n=15 and n+1 = 16
so it would be 15 (greek symbol) k=1 (1) = 15 (16)/2? @phi
  • phi
the sum of the numbers 1 to 15 is 15*16/2 = 15*8= 120 \[ \sum_{1}^{15}(2n+1)= 2\sum_{n=1}^{15}n+\sum_{n=1}^{15}1 \\ = 2(120) + \sum_{1}^{15}1 \] the sum of 15 ones is 15*1= 15 thus \[ \sum_{1}^{15}(2n+1)= 2\sum_{n=1}^{15}n+\sum_{n=1}^{15}1 \\ = 2(120) + 15 \] can you finish?
OMG ILY THANK YOU!!!!!!!!!!!

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