Can someone help with this? :)
It is posted below in the comment section so I can use the Equation editor.

- anonymous

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- anonymous

|dw:1362285707700:dw|\[\Sigma \]

- anonymous

Find the sum of the arithmetic series! Pleassse help:)

- hartnn

\(\huge \sum \limits_{k=1}^nk = \dfrac{n(n+1)}{2}\)
and a constant (here, 5) can be taken out of sum sign... :)

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## More answers

- anonymous

Where r we getting this formula? I'm confused where to start

- hartnn

thats a standard formula that can be used.

- hartnn

do you want a formula for sum of terms in arithmetic series ? its equivalent to this formula

- terenzreignz

Deriving the formula for an arithmetic series?

- anonymous

Yes that is the formula for the arithmetic series. Where do we go from there?:)

- anonymous

to find the sum?

- terenzreignz

Well, maybe best to use this formula instead
\[\huge \sum \limits_{k=1}^nk = \dfrac{n(n+1)}{2}\]
It's simpler :D

- terenzreignz

So anyway, this sum is just...
|dw:1362286375802:dw|
right?

- hartnn

ok, i'll mention the sum of arithmetic series formula also, then you choose which one you wanna use,
the series will be 5,10,15,20,.... right ? (with n= 100 terms)
he 1st term a1 = 5 and there's a common difference of d=5
then the sum formula is : \(S_n = (n/2)(2a+(n-1)d)\)

- anonymous

terenzreignz...how do you get these expanded form answer? Can u explain that? sorry

- anonymous

It is isn't the right answer in the back of the book

- terenzreignz

That's just kind of the definition. Or you could say the sigma notation is a compact form of the sum.
\[\huge \sum_{k=1}^n=1+2+3+...+(n-1)+n\]
Basically adding all integers starting at 1 to n.

- terenzreignz

\[\huge \sum_{k=1}^nk=1+2+3+...+(n-1)+n\]
Bloody typo, sorry :)

- terenzreignz

So anyway, we can rearrange...|dw:1362286994735:dw|

- terenzreignz

And notice that these terms...|dw:1362287037927:dw|
When added, they all equal n+1

- anonymous

what is the final answer? the answer is supposed to be 25,250...I just don't know how to get it

- hartnn

did you try any one of the 2 formula i gave u ??

- anonymous

Going back to hartnn's response near the beginning. This is like the only formua I'm understanding how to do.
\[S _{100}= (100(5 + ?))/2\]
I'm just confused how to get that number I didn't fill in? It's supposed to be \[t _{n}\] (which is last term)

- anonymous

Anyone?

- hartnn

oh, you need last term ?
the general term is 5k
to get last term, just put k=100 in general term
so, what u get as last term ?

- anonymous

So i fill in that question mark spot with 100?
Im not sure

- hartnn

general term =5k
with k=100, last term = 5*100 = 500
-_-

- anonymous

To clarify, this is the formula I am using.
\[S _{n}=\frac{ n(t _{1}+ t _{n} )}{ 2 }\]

- anonymous

wait how does k =100 again?

- hartnn

yup, thats the only other i didn't mention :P
because you are summing from k=1 (which is 1st term)
to k=100 (WHICH IS LAST TERM)

- hartnn

lower limit givs first term, upper limit gives last term

- anonymous

ok so 100 is the last term... And why do we multiply it by 5?

- hartnn

last value of k is 100
general term = 5k
so last term = 5*100

- anonymous

and the difference between the numbers would be five right? or is it something else

- hartnn

yes, common difference = 2nd term -1st term = 3rd term -2nd term =....
= 10-5 = 15-10 =.... = 5
but thats not required for the formula you are using

- anonymous

Oh ok that makes sense now. Maybe you should help me on other problems I post. Your a good help!

- hartnn

ok, sure, but did u get 25250 for this one ?

- anonymous

Oh wait quick question! What was \[t _{1}\] in this equation

- hartnn

t1 is the 1st term
index represents which term

- hartnn

tn is the n'th term

- anonymous

the index is which part again?

- hartnn

index is written as subscript
\(\huge t_n\)
here n is index and represent which term,
t1 = 1st term, t2 =2nd term, and so on

- anonymous

so t1 is 5

- hartnn

yup. as 5 is the 1st term

- hartnn

any more doubts ?

- anonymous

ok im sorry so I'm trying to solve it on paper. And I'm confused again.. how did we get \[t _{n}\] in the formula

- anonymous

oh wait my bad! i got it

- hartnn

you mean in \(S_n = n (t_1+t_n)/2\)
thats a general formula, do you want a derivation of this ?

- hartnn

oh u got it ? good :)

- anonymous

What would happen when I solve for \[t _{n}\] in an equation where n did not equal 1 and it equaled something else?

- hartnn

didn't get your question ? do u have specific example for that ?
say, if you want to find \(t_5\), then u put n=5 in general term tn

- anonymous

|dw:1362289579021:dw|
Like this. N does not equal 1

- hartnn

i think u have made up this question ? or is it from your book/notes ?

- anonymous

I made it up

- anonymous

here Ill give you one from the book.
|dw:1362289721775:dw|

- anonymous

thats 30- m by the way. It kinda got cut off

- hartnn

so, here your 1st term will start from m=10 in 30-m
and last term you'll get by putting m=20 in 30-m

- anonymous

so how would i get
\[t _{n}\]

- hartnn

tm = last term, last m =30
so to get last term tm, put m=30 in general term 30-m
is this confusing ?

- anonymous

kinda

- hartnn

general term tm =30-m
fist term t10 = 30-10 (put m=10)
last term t20 = 30-20 (put m=20)

- anonymous

so I would do 30 -10 first , and when do I do the 20? i dont know if that makes sense

- hartnn

it doesn't -_-

- anonymous

ok lets not do this problem.. Ill ask my teacher on monday..

- hartnn

you have formula, you have n, you have 1st term, u have last term....just plug in values!
ohh..ok, as u wish

- anonymous

Ill be posting other problems so u may help me if u want

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