Question

B??

Answer 1

I am not sure if I am doing this right but:

if you remove the sigma sign you just get the series 2k + 4

we replace k with n to get 2n + 4

@sweetburger may I have a second opinion on this?

Answer 2

@sweetburger

Answer 3

so to find a_n you are supposed to subtract Sn - S_(n-1)

Answer 4

okay

Answer 5

hmm let me think a bit on this one

Answer 6

Sure take your time

Answer 7

it's gotta be either B or D :s my math isn't working out well though

Answer 8

I'm gonna go to bed now because I am exhausted tomorrow we can get some other opinions if that's okay with you?

Answer 9

sure

Answer 10

sleep well

Answer 11

You as well =)

Answer 12

@sweetburger

Answer 13

@Shadow

Answer 14

@vannieduval @Vocaloid @katherine @ThisGirlPretty

Answer 15

@563blackghost

Answer 16

what do you guys think

Answer 17

@sillybilly123 @nnesha I am having a huge brain fart on this, I know the formula is an = Sn - S_n-1 but that's not giving me one of the choices

Answer 18

@SourMunchkin7806

Answer 19

i dont know this one im sorry did you try just saiyan

Answer 20

@JustSaiyan

Answer 21

@563blackghost

Answer 22

This is a hard one for me as well, I mean I have been trying to do this for about an hour now and I am just stuck. I would think it to be B based on `n` but my confidence in this answer is way low....

Answer 23

sorry :(

Answer 24

Ugh so like nobody knows I guess

Answer 25

\[\large\rm a_1= 2(1)+4=6\]
first term is 6
2nd term is 8
common difference =8-6=2
formula for nth term is \[\rm \large(common ~~difference)n+(first ~~~term -common~difference)\]

Answer 26

or this \[\rm a_1+(n-1)d\]

Answer 27

so C?

Answer 28

@Nnesha

Answer 29

no
\[\large\rm a_1= 2(1)+4=6\]
first term is 6
2nd term is 8
common difference =8-6=2
formula for nth term is
\[\rm \large(common ~~difference)n+(first ~~~term -common~difference)\]
so plugin the numbers \[\rm \large(2)n+(6-2)\]

Answer 30

^so B is correct.

Answer 31

Oh thanks =)

Answer 32

yw

Answer 33

n'th term would just be the nth partial sum which for any n must obey the same form of 2n+4.
SOrry for being 9 days late on this.