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Algebra 15 Online
OpenStudy (anonymous):

What are the first five partial sums of the geometric series: an = (5/8)*3^n-1?show work Student Answer: A. 0.625, 1.875, 5.625, 16.875, 50.625 B.1.875, 3.75, 7.5, 13.1250, 20.625 C. 2.5, 4.375, 10, 26.875, 77.5 d. 0.625, 2.5, 8.125, 25, 75.625

OpenStudy (anonymous):

Let's start by calculating n = 0.

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

is it a 11

OpenStudy (anonymous):

Wait... can you rewrite the problem with the equation tool?

OpenStudy (anonymous):

Just a_n

OpenStudy (anonymous):

\[a_n = \frac{ 5 }{ 8 }*3^n-1\] or\[a_n = \frac{ 5 }{ 8 }*3^{n-1}\]

OpenStudy (anonymous):

an(5/8)*3^n-1

OpenStudy (anonymous):

its the second one

OpenStudy (anonymous):

is it d

OpenStudy (anonymous):

but i need to show work ugh math

OpenStudy (anonymous):

Since it's the second one, we start with an index of 1. What is n = 1? Sorry for the lag in response.

OpenStudy (anonymous):

D is correct by the way. Let me know if you want an explanation.

OpenStudy (anonymous):

he told me i need to show my work can u help me plz

OpenStudy (anonymous):

Yes. Let's find the first partial sum. What is the sum when n = 1?

OpenStudy (anonymous):

um 4

OpenStudy (anonymous):

idk i am confused

OpenStudy (anonymous):

How did you get four? When n = 1, \[a_1 = (5/8)*3^0 = (5/8)*1 = 5/8\]

OpenStudy (anonymous):

oh my bad i wrote it wrong

OpenStudy (anonymous):

As in the series is written incorrectly?

OpenStudy (anonymous):

i put (5/8)*1+3 but i re written it

OpenStudy (anonymous):

Ok, so you understand how we got a_1 then?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

OK, this is our first partial sum. Now what is a_n when n =2? Adding a_1 and a_2 together will give you your second sum.

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

What did you get?

OpenStudy (anonymous):

so u add n-1+n-2

OpenStudy (anonymous):

No, my apologies for the confusing type. I mean,\[a_1 = 5/8\]\[a_2 = (5/8)*3^{1}\] Adding these together yields your second partial sum.

OpenStudy (anonymous):

Does that make sense?

OpenStudy (anonymous):

some what sorry i am bad at math

OpenStudy (anonymous):

i give up

OpenStudy (anonymous):

\[\sum_{n = 1}^{5}\frac{ 5 }{ 8 }3^{n-1} = \frac{ 5 }{ 8 }3^0 + \frac{ 5 }{ 8 }3^1 +\frac{ 5 }{ 8 }3^2 +...+\frac{ 5 }{ 8 }3^5 = \frac{ 5 }{ 8 }(1 + 3 + 9 + 27 + 81)\] \[\sum_{n = 1}^{1}\frac{ 5 }{ 8 }3^{n-1} = \frac{ 5 }{ 8 }3^0\] \[\sum_{n = 1}^{2}\frac{ 5 }{ 8 }3^{n-1} = \frac{ 5 }{ 8 }3^0 + \frac{ 5 }{ 8 }3^1\] . . .

OpenStudy (anonymous):

then u divide right

OpenStudy (anonymous):

Divide? What would you divide by?

OpenStudy (anonymous):

ugh idk this is hard

OpenStudy (anonymous):

I'm sorry that I cannot help you better.

OpenStudy (anonymous):

It's really quite simple. Maybe try again later?

OpenStudy (anonymous):

i dont know what to do on this is hard for me lol

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