anonymous
  • anonymous
i use the formula, but don't understand it.... (sum of finite geometric series)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1435876640907:dw|
anonymous
  • anonymous
@freckles
freckles
  • freckles
do you mean how do you derive that to be the formula?

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anonymous
  • anonymous
i mean how do I understand it....
anonymous
  • anonymous
Every formula I have previously used I understand why it works to calculate a certain thing....
anonymous
  • anonymous
but, this....
freckles
  • freckles
Ok well let me derive it for you and maybe you will understand it better... \[a_1,ra_1,r^2a_1,r^3a_1..., \text{ is a geometric sequence } \\a_2=ra_1 \\ a_3=r^2a_1 \\ \cdots \\ a_n=r^{n-1}a_1 \\ \ \\ \text{ geometric series is the sum of the terms of the geometric sequence }\] \[S_n=a_1+ra_1+r^2a_1+r^3a_1 \cdots +r^{n-1}a_1 \\S_n= a_1(1+r+r^2+r^3+ \cdots r^{n-1}) \\ \] ...
freckles
  • freckles
ok before I go on... do you know what we get when we do: \[\frac{r^n-1}{r-1}\]
anonymous
  • anonymous
when n->∞ ? or what do you mean?
freckles
  • freckles
examples: \[\frac{r^2-1}{r-1}=r+1 \\ \frac{r^3-1}{r-1}=r^2+r+1 \\ \frac{r^4-1}{r-1}=(r+1)(r^2+1)=r^3+r^2+r+1 \\ \frac{r^5-1}{r-1}=r^4+r^3+r^2+r+1\] so on...
freckles
  • freckles
basically the trick here to recognize \[r^{n-1}+r^{n-2}+r^{n-1}+ \cdots +r^{3}+r^2+r+1=\frac{r^n-1}{r-1}\]
anonymous
  • anonymous
oh, so you get r^(n-1)+r^(n-2)+....r^3+r^2+r+1
anonymous
  • anonymous
yeah
freckles
  • freckles
oops I didn't mean to write n-1 again
anonymous
  • anonymous
yes I see, no need correction
freckles
  • freckles
\[r^{n-1}+r^{n-2}+r^{n-\color{red}3}+ \cdots +r^{3}+r^2+r+1=\frac{r^n-1}{r-1}\]
freckles
  • freckles
ok I did it anyways :p
anonymous
  • anonymous
so, \(\LARGE \frac{r^{n}-1}{r-1}\) is just the sum of all these r^(n-1)+....+r+1 but these aren't terms, because the terms are when each of these is multiplying times a1, so..... a1 • r^(n-1) + a1 • r^(n-2) + .... a1•r+a1•1
freckles
  • freckles
\[S_n=a_1+ra_1+r^2a_1+r^3a_1 \cdots +r^{n-1}a_1 \\S_n= a_1(1+r+r^2+r^3+ \cdots r^{n-1}) \\ \\ S_n=a_1 \frac{r^n-1}{r-1} \\ \text{ or multiply both \top and bottom gives } S_n=a_1 \frac{1-r^n}{1-r}\]
anonymous
  • anonymous
the trick is to notice that if we let \(S_n\) denote the sum of \(n\) terms we get: $$S_n=a_1+a_1 r+a_1 r^2+\dots+a_1r^{n-1}\\r S_n=a_1 r+a_1 r^2+\dots+a_1r^{n-1}+a_1 r^n\\S_n-rS_n=a_1-a_1 r^n\\(1-r)S_n=a_1(1-r^n)\\S_n=a_1\frac{1-r^n}{1-r}$$
freckles
  • freckles
yes you are right so you can take that sum of the r things you wrote and multiply it by your initial
freckles
  • freckles
that's cute @oldrin.bataku
anonymous
  • anonymous
So, this is what we do: we take the (1-r^(n-1)) / (1-r) that = r^(n-1) + r^(n-2) + .... 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.
freckles
  • freckles
yep
anonymous
  • anonymous
tnx 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
  • anonymous
You really saved me. "Filled in the gap"
freckles
  • freckles
np didn't do much :)
anonymous
  • anonymous
yeah 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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