Question

evaluate 10 sigma n=2 -2n-3? how do you do this

Answer 1

Do the same thing I showed you earlier \[\sum_{n=2}^{10} -2n-3 \implies -2(2)-3+-2(3)-3+...\] start at n = 2 and end at n = 10

Answer 2

what is the + is for after 3?

Answer 3

That means the pattern continues, so do it up till 10

Answer 4

All you're doing is plugging in the "n's" into the equation

Answer 5

and sigma implies to sum it up, so you plug in n = 2 and keep plugging in n's up till 10, and then you add them all up

Answer 6

Is -135 the correct answer. that what I got when I add them all up

Answer 7

Good job!

Answer 8

evaluate 5 \[\sum_{n=1}^{5} 3(-2)^{n-1}\]

Answer 9

Same thing as before except now n is a exponent

Answer 10

Remember an exponent tells you how many times you multiply by itself and anything to the power of 0 = 1, \[x^0=1\]

Answer 11

so is it 3(-2)^1-1
3(-2)^2-1

Answer 12

yes exactly up to 5

Answer 13

I feel like I did something wrong because i add them up it is equal to -71

Answer 14

my options are
a.−93
b.−33
c.33
d.93

Answer 15

\[3(-2)^{1-1} = 3(-2)^0 = 3(1) = 3\]

Answer 16

\[3(-2)^{2-1} = 3(-2)^1 = -6\]

Answer 17

\[3(-2)^{3-1} =3(-2)^2 = 3(4)=12\]

Answer 18

now do n = 4 and n = 5

Answer 19

\[3(-2)^{4-1} = 3(-2)^{3} =?\]

Answer 20

@Astrophysics

Answer 21

@Ashleyisakitty

Answer 22

@Agl202

Answer 23

got it thank you very much

Answer 24

\[\sum_{n=3}^{12} 20(0.5)^{n-1}\]