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Algebra 14 Online
OpenStudy (allieeslabae):

Help?

OpenStudy (allieeslabae):

OpenStudy (allieeslabae):

OpenStudy (thomas5267):

For the first one, it could be a or d. I don't know what n starts on so I don't know.

OpenStudy (allieeslabae):

N is the numbers they provide like -2... thats a1

OpenStudy (allieeslabae):

I was thinking A or D as well. I'm torn because we can only choose one, and only one is correct.

OpenStudy (thomas5267):

If the index starts at 0, i.e. \(a_0,a_1,a_2,\dots\), then it is pretty clear that it is a. If the index starts at 1, i.e. \(a_1,a_2,a_3,\dots\), then it is pretty clear that it is d. b and c does not make sense since if you increase n by one \(a_n\) would also increase by one. In other words, for b and c, \(a_{n+1}-a_n=1\).

OpenStudy (allieeslabae):

Okay so you're thinking its A? Okay cool how about the next two?

OpenStudy (allieeslabae):

@thomas5267

OpenStudy (thomas5267):

How do you write down a sequence? Does the index start from 0 or 1?

OpenStudy (allieeslabae):

It starts with 1.

OpenStudy (thomas5267):

For the write down the explicit definition for the sequence question it should be D.

OpenStudy (allieeslabae):

Yeperonni! thank you!

OpenStudy (thomas5267):

For the sum of first 10 terms question, use the formula for arithmetic series.

OpenStudy (allieeslabae):

I got C how does that sound?

OpenStudy (thomas5267):

You missed the factor of 2?

OpenStudy (allieeslabae):

Hang on.

OpenStudy (allieeslabae):

Okay im confused. How do you set this up?

OpenStudy (thomas5267):

Using \(\displaystyle S_n=\frac{n(a_1+a_n)}{2}\).

OpenStudy (allieeslabae):

S10=10(41+an)/2 ???

OpenStudy (thomas5267):

That's the general formula. In this case, \(\displaystyle S_{10}=\frac{10(a_1+a_{10})}{2}\)

OpenStudy (allieeslabae):

Ahh okay so B?

OpenStudy (thomas5267):

Yep.

OpenStudy (allieeslabae):

mm wonder why its not D or C?

OpenStudy (thomas5267):

Just add the 10 terms manually and you will understand why it is not C or D. XD

OpenStudy (allieeslabae):

AHAHAH oh wow.. missed the sum part. Okay thanks!

OpenStudy (thomas5267):

For the geometric series, use the formula again. \[ \sum_{k=0}^nar^k=a\frac{1-r^{n+1}}{1-r} \] Note that the sum runs from k=0 to k=n and the it is \(r^{n+1}\) on the numerator of RHS.

OpenStudy (allieeslabae):

I noticed that 0 which makes it 0 right?

OpenStudy (thomas5267):

???

OpenStudy (allieeslabae):

k-0

OpenStudy (allieeslabae):

Let me just use your formula instead of my teachers.. hang on.

OpenStudy (thomas5267):

No. \(\displaystyle \sum_{k=0}^nar^k\) means \(ar^0+ar^1+ar^2+\cdots\) all the way up to n.

OpenStudy (allieeslabae):

I seriously dont see how you got that. Im still getting 4.

OpenStudy (thomas5267):

Yes the answer is 4. Now I am confused on how you got the answer.

OpenStudy (allieeslabae):

BWHAHA thank you anyways!

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