jmartinez638
  • jmartinez638
Calculate the following series:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jmartinez638
  • jmartinez638
\[\sum_{k=1}^{20} 3k^2 +5k\]
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
break it apart first

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misty1212
  • misty1212
\[3\sum k^2+5\sum k\] then use the formulas for each
jmartinez638
  • jmartinez638
\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (4+2k/3\times1/n)\times1/n\]
jmartinez638
  • jmartinez638
Hiya! ;)
misty1212
  • misty1212
i can't read the second one
jmartinez638
  • jmartinez638
\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (1+2k/n)^8 \times1/n\]
jmartinez638
  • jmartinez638
One sec, I will edit it.
misty1212
  • misty1212
\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (4+\frac{2k}{3})\times\frac{1}{n})\times\frac{1}{n}\] is my guess
misty1212
  • misty1212
some sorta integral right?
misty1212
  • misty1212
turn it to \[\frac{1}{n^2}\sum 4+\frac{2}{3}\sum k\]
misty1212
  • misty1212
actually \[\frac{1}{n^2}\left(\sum_{k=1}^n 4+\frac{2}{3}\sum_{k=1}^n k\right)\]
misty1212
  • misty1212
unless am reading it wrong, which is possible
Zarkon
  • Zarkon
\[\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (1+2k/n)^8 \times1/n\] are you supposed to write this as an integral and evaluate?
jmartinez638
  • jmartinez638
Just says evaluate, your thought is as good as mine.
jmartinez638
  • jmartinez638
Not to be blunt, I just don't know

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