Mathematics
10 Online
OpenStudy (anonymous):
does anyone know what the general terms of this sequence is 3,6,10,15
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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
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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?
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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)
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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?
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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
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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