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Mathematics 10 Online
OpenStudy (anonymous):

does anyone know what the general terms of this sequence is 3,6,10,15

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

lol can you help me

OpenStudy (anonymous):

sorry i didnt got you

OpenStudy (anonymous):

it is a arithmetic sequence with a twist.

OpenStudy (anonymous):

so how do i do it

OpenStudy (anonymous):

how does each sequence differs

OpenStudy (anonymous):

it increase by one each time so ,2,3,4,5

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

now what do i do

OpenStudy (anonymous):

do you know the form for an arithmetic sequence?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

ok ill give you the form for that and then you can try to relate it to this problem. \[\sum_{n=1}^{N}n=N*(N+1)/2\]

OpenStudy (anonymous):

let me know if that's enough to get you started

OpenStudy (anonymous):

what does n represent

OpenStudy (anonymous):

n is just an integer. in the sum if you have N=3 you get 1+2+3=6= (3*4/2)

OpenStudy (anonymous):

can you these formulas for series?

OpenStudy (anonymous):

yes it will help you find a general form for this series in terms of n where n represents with term in the series you are at.

OpenStudy (anonymous):

do we alway use this formula for finding the general term

OpenStudy (anonymous):

no not always

OpenStudy (anonymous):

are you looking for the series(sum) or sequence?

OpenStudy (anonymous):

for the series

OpenStudy (anonymous):

i just want a quick way of figuring out the general term

OpenStudy (anonymous):

instead of keep guessing

OpenStudy (anonymous):

lets break the terms into sums 1+2 1+2+3 1+2+3+4 ... so if n denotes the term we have \[\sum_{i=1}^{n+1}i=(n+1)*(n+2)/2=(n^{2}+3n+2)/2\]

OpenStudy (anonymous):

so n=1: (1+3+2)/2=3 you can check the other terms if you want

OpenStudy (anonymous):

is there a a better way of figuring this out so you can avoid fractions in the general term

OpenStudy (anonymous):

this is the simplest way I know to do it.

OpenStudy (anonymous):

okay ty

OpenStudy (anonymous):

yep, hope it helped a bit :)

OpenStudy (anonymous):

yah it did ty

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