Question

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.

Answer 1

https://gyazo.com/838e3b821bb0bb6c5de207edcd36eae4

Answer 2

@jim_thompson5910 Would there be no shortcut? Other adding each one.

Answer 3

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

Answer 4

and also

\[\Large \sum_{i = 1}^{n}k = k*n\]

where k is some constant

Answer 5

\[\Large \sum_{i=1}^{10}\left(2i^2+4i-5\right) = \sum_{i=1}^{10}\left(2i^2\right)+\sum_{i=1}^{10}\left(4i\right)+\sum_{i=1}^{10}\left(-5\right)\]


\[\Large \sum_{i=1}^{10}\left(2i^2+4i-5\right) = 2*\sum_{i=1}^{10}\left(i^2\right)+4*\sum_{i=1}^{10}\left(i\right)+\sum_{i=1}^{10}\left(-5\right)\]


I'll let you finish

Answer 6

I don't know what I did wrong I got 985

Answer 7

Tell me what you got for
\[\Large \sum_{i=1}^{10}\left(i^2\right)\]
and
\[\Large \sum_{i=1}^{10}\left(i\right)\]

Answer 8

After multiplying by two I got 770 for the first one

Answer 9

2nd one I got 220 after multiplying by 4

Answer 10

\[\Large \sum_{i=1}^{10}\left(i^2\right)\] is not equal to 770

Answer 11

oh you mean \[\Large 2*\sum_{i=1}^{10}\left(i^2\right)\] ???

Answer 12

Yes

Answer 13

\[\Large 2*\sum_{i=1}^{10}\left(i^2\right) = 770\] is correct

Answer 14

\[\Large 4*\sum_{i=1}^{10}\left(i\right) = 4*55 = 220\] is also correct

Answer 15

what is \[\Large \sum_{i=1}^{10}\left(-5\right)\] equal to?

Answer 16

Won't it just be -5?

Answer 17

nope

Answer 18

think of it as you're summing 10 copies of (-5)

Answer 19

-50

Answer 20

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

Answer 21

as a shortcut, you can say 10*(-5) = -50, so yes you have it

Answer 22

Thank you good sir

Answer 23

I have one final question I'd appreciate it if you could look over it.