Question

summation notation

Answer 1

Write the sum using summation notation, assuming the suggested pattern continues.

-9 - 3 + 3 + 9 + ... + 81

Answer 2

looks like you are adding \(6\) each time

Answer 3

you could try
\[\sum_{k=0}^{30}-9+3k\]

Answer 4

or if you want to start at \(k=1\) instead of \(k=0\) you could use
\[\sum_{k=1}^{31}-12+3k\]

Answer 5

wait what difference does that make?

Answer 6

ok first of all i must be on crack
i said "add 6" each time not "add 3"

Answer 7

\[\sum_{k=0}^{15}-9+6k\]

Answer 8

lol indeed, you said add 6 XD

Answer 9

so my first answer was silly
go with the second one
either that, or
\[\sum_{k=1}^{16}-12+6k\]will do it

Answer 10

wouldnt it be 15 on top?

Answer 11

if you are wondering why i started at 1, that is because frequently you start the summation at \(k=1\) rather than \(k=0\)

Answer 12

but if \(k=1\) is where you start then you need \(-15+6k\) so that when \(k=1\) you get the first term as
\[-15+6\times 1=-9\]

Answer 13

well it has to be n=0 cuz thats the only thing in my answers

Answer 14

ok
to figure out where you end, then solve
\[-9+6k=81\] for \(k\) and you get \(k=15\) (i hope) which is where you stop

Answer 15

sorry about my early mistake, it was stupid

Answer 16

okay so i was right 15 goes ontop

Answer 17

yes

Answer 18

your questions are all over the place
complex numbers
vectors
sigma notation
precalculus on line maybe?

Answer 19

so its 15Ethingy.....n=0 and then beside the e thingy is (-9+6n)

Answer 20

yes
E thingy called "sigma"

Answer 21

means add

Answer 22

\[\sum_{n=0}^{15}-9+6n\]

Answer 23

yeah XD pre calc tends to do that a girl told me a few days ago that youre suppose to evaluate limits before you learn about derivitives.... and my online book does it opposite :p the online books a stupid, they dont even help any its like reading gibberish such as sigma ^-^

Answer 24

yeah it is tough to read this notation if you are not used to it
it is very helpful later because no one want to write
\[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\] when you can just write
\[\sum_{n=1}^{\infty}\frac{1}{2^n}\]

Answer 25

derivatives and limits are calculus, not precalc
having to read the book on your own must be a pain

Answer 26

i literaly had just asked a question earlier that asked something about 1/2, 1/2^2, 1/2^3, 1/2^4, 1/2^5