anonymous
  • anonymous
Use the properties of sigma notation and the summation formulas to evaluate the summation from i equals 1 to 10 of the quantity 2 times i squared plus 4 times i minus 5.
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
https://gyazo.com/838e3b821bb0bb6c5de207edcd36eae4
anonymous
  • anonymous
@jim_thompson5910 Would there be no shortcut? Other adding each one.
jim_thompson5910
  • jim_thompson5910
you'll use these properties http://www.phengkimving.com/calc_of_one_real_var/09_the_intgrl/09_01_summ_notatn_and_formulas_files/image012.gif specifically property i and property ii

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jim_thompson5910
  • jim_thompson5910
and also \[\Large \sum_{i = 1}^{n}k = k*n\] where k is some constant
jim_thompson5910
  • jim_thompson5910
\[\Large \sum_{i=1}^{10}\left(2i^2+4i-5\right) = \sum_{i=1}^{10}\left(2i^2\right)+\sum_{i=1}^{10}\left(4i\right)+\sum_{i=1}^{10}\left(-5\right)\] \[\Large \sum_{i=1}^{10}\left(2i^2+4i-5\right) = 2*\sum_{i=1}^{10}\left(i^2\right)+4*\sum_{i=1}^{10}\left(i\right)+\sum_{i=1}^{10}\left(-5\right)\] I'll let you finish
anonymous
  • anonymous
I don't know what I did wrong I got 985
jim_thompson5910
  • jim_thompson5910
Tell me what you got for \[\Large \sum_{i=1}^{10}\left(i^2\right)\] and \[\Large \sum_{i=1}^{10}\left(i\right)\]
anonymous
  • anonymous
After multiplying by two I got 770 for the first one
anonymous
  • anonymous
2nd one I got 220 after multiplying by 4
jim_thompson5910
  • jim_thompson5910
\[\Large \sum_{i=1}^{10}\left(i^2\right)\] is not equal to 770
jim_thompson5910
  • jim_thompson5910
oh you mean \[\Large 2*\sum_{i=1}^{10}\left(i^2\right)\] ???
anonymous
  • anonymous
Yes
jim_thompson5910
  • jim_thompson5910
\[\Large 2*\sum_{i=1}^{10}\left(i^2\right) = 770\] is correct
jim_thompson5910
  • jim_thompson5910
\[\Large 4*\sum_{i=1}^{10}\left(i\right) = 4*55 = 220\] is also correct
jim_thompson5910
  • jim_thompson5910
what is \[\Large \sum_{i=1}^{10}\left(-5\right)\] equal to?
anonymous
  • anonymous
Won't it just be -5?
jim_thompson5910
  • jim_thompson5910
nope
jim_thompson5910
  • jim_thompson5910
think of it as you're summing 10 copies of (-5)
anonymous
  • anonymous
-50
jim_thompson5910
  • jim_thompson5910
smaller example \[\Large \sum_{i=1}^{4}\left(-5\right) = \left(-5\right)+\left(-5\right)+\left(-5\right)+\left(-5\right)\] notice how there are 4 copies of (-5) being added
jim_thompson5910
  • jim_thompson5910
as a shortcut, you can say 10*(-5) = -50, so yes you have it
anonymous
  • anonymous
Thank you good sir
anonymous
  • anonymous
I have one final question I'd appreciate it if you could look over it.

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