Question

http://openstudy.com/study#/updates/520eac21e4b00ebc7dbeaf09 can anyone help me the problem is in the link.

Answer 1

oh god...

Answer 2

Micro can you help?

Answer 3

Sure

Answer 4

omg thx lol see that wasnt so hard XD

Answer 5

Ok, to solve this type of question

Answer 6

Do you see the sigma notation sign, the middle part is basically the function

Answer 7

i=1 is the number you will start with

Answer 8

7 is the one you stop right at

Answer 9

Yes, i see it, and i think i know how to do it but im not sure about the i=1 thing under the sigma. usually its an n so i dont know if the i affects anything

Answer 10

Ok the i=1 is the number you start with

Answer 11

I have problems with complex numbers and such so yea

Answer 12

There is no difference between an 'n' and a 'i'???

Answer 13

and the number on top is the number you stop at so in your case it is 7

Answer 14

Ok now your function is 2(3)^(i-1)

Answer 15

so i substitute the numbers 1-7 into the function and thats my answer?

Answer 16

Yeah

Answer 17

Ok, how do i know if its convergent or divergent?

Answer 18

Ok to find that out

Answer 19

Give me one second

Answer 20

Ok np

Answer 21

wait...brainfart.....in the equation do i multiply the 2 times the 3 first or do i do the exponent

Answer 22

You do the exponent first

Answer 23

For example 2-1 for the exponent gives me 1

Answer 24

2*3= 6^1=6

Answer 25

This is convergent

Answer 26

Because it has a limit

Answer 27

Which is 7

Answer 28

Any questions?

Answer 29

No this is perfect sorry i was writing and taking notes thank you so much
A MEDAL FOR JU.

Answer 30

Thank you

Answer 31

no problem *fan*
ur a genius sent from the heavens.

Answer 32

Thank you, have a good day