Write the sum using summation notation, assuming the suggested pattern continues. -10 - 2 + 6 + 14 + ... + 110
So the equation for a finite arithmetic sum is \[S_{n} =\frac{ n }{ 2 }(a_1 + a_n)\]All we need to do is find n. So we have \[a_n = a_1 + (n-1)d\]And since a_1 is -10 and a_n is 110, and d is obviously 8, we have \[n=\frac{ 110-(-10) }{ 8 } + 1\] Which is 16. So our sum is\[\frac{ 16 }{ 2 }(-10 + 110) = 8(100) = 800\]Good?
Write the sum using summation notation: \[\Large \sum_{n = 1}^{16}(8n - 18)\]
Oh, summation notation. Sorry. That's right.
Wait how did you get all thst
All what?
no what he wrote
I was basing my answer on OakTree's calculations above.
I think she means explain where the argument of the summation came from.
We need to find the nth term of the arithmetic series.
This is an arithmetic series with first term -10 and common difference 8. The nth term is given by a_n = a_1 + (n-1)d = -10 + (n-1)(8) = -10 + 8n - 8 a_n = 8n - 18. We need to sum this as n runs from 1 to 16 (16 was determined to be the number of terms in the series by OakTree above).
You can factor the 2 out of the summation.
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