Help would be appreciated!
Please explain how you got the answer c:

Answer:
-25
Step-by-step explanation:
We're given the sigma notation of a series.
[tex] \mathsf{ \sum _{k = 1} ^{5}(2k - 11) }[/tex]
The variable k is the "index of notation"
The expression can be read as the sum of (2k- 11) as k goes from 1 to 5.
"To generate the terms of a series given in sigma notation, successively replace the index of summation with consecutive integers from the first value to the last value of the index" ~internet
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To generate the terms of the series given on sigma notation above, replace k by 1, 2, 3, 4, and 5 and add the terms, to get the answer to the given notation.
[tex]\mathsf{ \sum _{k = 1} ^{5}(2k - 11) } \\ = \mathsf{ \overline{2(1) - 11} + \overline{2(2) - 11} + \overline{2(3) - 11} } \\ \mathsf{ + \overline{2(4) - 11} + \overline{2(5) - 11}}[/tex]
[tex] = \mathsf{\overline{2 - 11} +\overline{4 - 11} + \overline{6 - 11} } \\ \mathsf{\overline{8 - 11} + \overline{10- 11}}[/tex]
[tex] = \mathsf{ - 9- 7 - 5 - 3 - 1}[/tex]
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[tex] \mathsf{ \underline{- 25}}[/tex]
______________________
[tex] \mathfrak{ \overline{There \: You \: Are!}}[/tex]
Let's see
[tex]\\ \rm\Rrightarrow {\displaystyle{\sum^2_{k=1}}}(2k-11)[/tex]
[tex]\\ \rm\Rrightarrow 2(1)-11+2(2)-11+2(3)-11+2(4)-11+2(5)-11[/tex]
[tex]\\ \rm\Rrightarrow 2-11+4-11+6-11+8-11+10-11[/tex]
[tex]\\ \rm\Rrightarrow 30-55[/tex]
[tex]\\ \rm\Rrightarrow -25[/tex]