Answered

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Evaluate:
[tex]\bf{\sum^{32}_{n=1}\:(2n+8)[/tex]

Please help A.S.A.P., thank you guys!

Please show work as well.

[tex]\bigstar[/tex]

Sagot :

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Sā‚ƒā‚‚ = 1312

Explanation:

[tex]\boxed{\sf \sum _{n=1}^{32}\left(2n+8\right)}[/tex]

Formula Required:

[tex]\sf sum \ of \ arithmetic \ series = \dfrac{n}{2} (2a \ + \ (n-1) \ d)[/tex]

Identify the following's:

  • First Term (a) = (2(1) + 8) = 10

  • Common Difference (d) = 2nd term - first term = (2(2) + 8) - (2(1) + 8) = 2

  • Total Terms (n) = 32

Insert variables:

[tex]\rightarrow \ \sf S_{32} = \dfrac{32}{2} (2(10) + (32-1) 2) \quad \xrightarrow{\text{Simplify} } \quad 1312[/tex]

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