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
Let's start by calculating n = 0.
okay
is it a 11
Wait... can you rewrite the problem with the equation tool?
Just a_n
\[a_n = \frac{ 5 }{ 8 }*3^n-1\] or\[a_n = \frac{ 5 }{ 8 }*3^{n-1}\]
an(5/8)*3^n-1
its the second one
is it d
but i need to show work ugh math
Since it's the second one, we start with an index of 1. What is n = 1? Sorry for the lag in response.
D is correct by the way. Let me know if you want an explanation.
he told me i need to show my work can u help me plz
Yes. Let's find the first partial sum. What is the sum when n = 1?
um 4
idk i am confused
How did you get four? When n = 1, \[a_1 = (5/8)*3^0 = (5/8)*1 = 5/8\]
oh my bad i wrote it wrong
As in the series is written incorrectly?
i put (5/8)*1+3 but i re written it
Ok, so you understand how we got a_1 then?
yeah
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.
okay
What did you get?
so u add n-1+n-2
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.
Does that make sense?
some what sorry i am bad at math
i give up
\[\sum_{n = 1}^{5}\frac{ 5 }{ 8 }3^{n-1} = \frac{ 5 }{ 8 }3^0 + \frac{ 5 }{ 8 }3^1 +\frac{ 5 }{ 8 }3^2 +...+\frac{ 5 }{ 8 }3^5 = \frac{ 5 }{ 8 }(1 + 3 + 9 + 27 + 81)\] \[\sum_{n = 1}^{1}\frac{ 5 }{ 8 }3^{n-1} = \frac{ 5 }{ 8 }3^0\] \[\sum_{n = 1}^{2}\frac{ 5 }{ 8 }3^{n-1} = \frac{ 5 }{ 8 }3^0 + \frac{ 5 }{ 8 }3^1\] . . .
then u divide right
Divide? What would you divide by?
ugh idk this is hard
I'm sorry that I cannot help you better.
It's really quite simple. Maybe try again later?
i dont know what to do on this is hard for me lol
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