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anonymous
 one year ago
i use the formula, but don't understand it....
(sum of finite geometric series)
anonymous
 one year ago
i use the formula, but don't understand it.... (sum of finite geometric series)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435876640907:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you mean how do you derive that to be the formula?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean how do I understand it....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Every formula I have previously used I understand why it works to calculate a certain thing....

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Ok well let me derive it for you and maybe you will understand it better... \[a_1,ra_1,r^2a_1,r^3a_1..., \text{ is a geometric sequence } \\a_2=ra_1 \\ a_3=r^2a_1 \\ \cdots \\ a_n=r^{n1}a_1 \\ \ \\ \text{ geometric series is the sum of the terms of the geometric sequence }\] \[S_n=a_1+ra_1+r^2a_1+r^3a_1 \cdots +r^{n1}a_1 \\S_n= a_1(1+r+r^2+r^3+ \cdots r^{n1}) \\ \] ...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok before I go on... do you know what we get when we do: \[\frac{r^n1}{r1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when n>∞ ? or what do you mean?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1examples: \[\frac{r^21}{r1}=r+1 \\ \frac{r^31}{r1}=r^2+r+1 \\ \frac{r^41}{r1}=(r+1)(r^2+1)=r^3+r^2+r+1 \\ \frac{r^51}{r1}=r^4+r^3+r^2+r+1\] so on...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1basically the trick here to recognize \[r^{n1}+r^{n2}+r^{n1}+ \cdots +r^{3}+r^2+r+1=\frac{r^n1}{r1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, so you get r^(n1)+r^(n2)+....r^3+r^2+r+1

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oops I didn't mean to write n1 again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes I see, no need correction

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[r^{n1}+r^{n2}+r^{n\color{red}3}+ \cdots +r^{3}+r^2+r+1=\frac{r^n1}{r1}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok I did it anyways :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, \(\LARGE \frac{r^{n}1}{r1}\) is just the sum of all these r^(n1)+....+r+1 but these aren't terms, because the terms are when each of these is multiplying times a1, so..... a1 • r^(n1) + a1 • r^(n2) + .... a1•r+a1•1

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[S_n=a_1+ra_1+r^2a_1+r^3a_1 \cdots +r^{n1}a_1 \\S_n= a_1(1+r+r^2+r^3+ \cdots r^{n1}) \\ \\ S_n=a_1 \frac{r^n1}{r1} \\ \text{ or multiply both \top and bottom gives } S_n=a_1 \frac{1r^n}{1r}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the trick is to notice that if we let \(S_n\) denote the sum of \(n\) terms we get: $$S_n=a_1+a_1 r+a_1 r^2+\dots+a_1r^{n1}\\r S_n=a_1 r+a_1 r^2+\dots+a_1r^{n1}+a_1 r^n\\S_nrS_n=a_1a_1 r^n\\(1r)S_n=a_1(1r^n)\\S_n=a_1\frac{1r^n}{1r}$$

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yes you are right so you can take that sum of the r things you wrote and multiply it by your initial

freckles
 one year ago
Best ResponseYou've already chosen the best response.1that's cute @oldrin.bataku

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, this is what we do: we take the (1r^(n1)) / (1r) that = r^(n1) + r^(n2) + .... r + 1 and then so that we are going to be adding the terms, we multiply by the a1 component: And this way we get the formula that is there.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tnx again, you got my back I will say I don't know jack. when compare to you at least, because you are a math beast.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You really saved me. "Filled in the gap"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah you did. YOu kinda made my knowledge a whole piece. A quite small piece, but a whole one.... in any case.... cu, and ty very much
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