Question

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

-1 + 2 + 5 + 8 + ... + 44

Answer 1

so whats our pattern?

Answer 2

+3 I'm mainly confused on how to set up the summation notation.

Answer 3

@amistre64

Answer 4

well if we have a rule, then we can just tell the summation to add up all the terms defined by the rule

we start at -1: and add 3 soo

-1 +3n is a good rule, for n=0,1,2,3,...

Answer 5

okay

Answer 6

if we want to sum up all the terms, \[\Sigma\text{; is just a capital S, for 'summation"}\]

\[\sum_{n=low}^{high}rule\]

Answer 7

instead of greek :)

\[\{Sum\}_{n=0}^{n=k}~~~(rule)\]

Answer 8

sooo \[\sum_{n=0}^{15} (-1+3n) ?????\] or would it be infinity instead of 15..??

Answer 9

hmm, i spose we should try to figure out how many terms we have ...

Answer 10

-1 + 3n = 44

what is n?

Answer 11

45/3 = 15 yeah, thats fine :)

Answer 12

okay ^.^

Answer 13

so I'm right??

Answer 14

first term: n=0, we get -1

last term: n=15, we get 44

thats all of them

Answer 15

yep, you did great

Answer 16

yay! :D

Answer 17

can you help me with more?? i suck at this stuff... >.<

Answer 18

one more, maybe

Answer 19

thanks i'll make it a good one lol

Answer 20

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

4 - 24 + 144 - 864 + ...

Answer 21

alternating signs eh, whats that suggest?

Answer 22

multiplication is my best guess....

Answer 23

yep, and (-1)^n so that the signs alternate

now what our ratio?

Answer 24

-6...? maybe.. i'm not confident in this answer...

Answer 25

your right, dont worry

(-6)^n = (-1)^n 6^n

so it all depends on how you want to present that part of it

Answer 26

then n=0, (-6)^n = 1 and we multiply our first term

so whats our rule?

Answer 27

consider

4 = a (-6)^0

what is a?

Answer 28

the rule will always be the first term plus the ratio right??

Answer 29

well, times the ratio in this case. its common differences that add, ratios multiply

Answer 30

a0 = a
a1 = ar
a2 = arr
a3 = arrr


each new term is the last one times r

Answer 31

that confused me more...

Answer 32

4(-6)^n-1??

Answer 33

does n start at 1 or 0?

Answer 34

i'm not sure.... but im guessing 0 so it would be...
4(-6)^n

Answer 35

its author dependant, i start at n=0, some courses say n=1 so there is really no standard to go by.

as long as we define n starts at 0 it makes the rule simpler to write

Answer 36

now we hav eno final term, so what our upper limit?

Answer 37

so, write up the summation notation :)

Answer 38

\[\sum_{n=0}^{\infty} 4(-6)^n \] ...?

Answer 39

perfect

Answer 40

YAY :D Thank you soooooooooo much!!!!!!

Answer 41

good luck :)

Answer 42

thanks! :)