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

1. anonymous

|dw:1435876640907:dw|

2. anonymous

@freckles

3. freckles

do you mean how do you derive that to be the formula?

4. anonymous

i mean how do I understand it....

5. anonymous

Every formula I have previously used I understand why it works to calculate a certain thing....

6. anonymous

but, this....

7. freckles

Ok well let me derive it for you and maybe you will understand it better... $a_1,ra_1,r^2a_1,r^3a_1..., \text{ is a geometric sequence } \\a_2=ra_1 \\ a_3=r^2a_1 \\ \cdots \\ a_n=r^{n-1}a_1 \\ \ \\ \text{ geometric series is the sum of the terms of the geometric sequence }$ $S_n=a_1+ra_1+r^2a_1+r^3a_1 \cdots +r^{n-1}a_1 \\S_n= a_1(1+r+r^2+r^3+ \cdots r^{n-1}) \\$ ...

8. freckles

ok before I go on... do you know what we get when we do: $\frac{r^n-1}{r-1}$

9. anonymous

when n->∞ ? or what do you mean?

10. 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...

11. freckles

basically the trick here to recognize $r^{n-1}+r^{n-2}+r^{n-1}+ \cdots +r^{3}+r^2+r+1=\frac{r^n-1}{r-1}$

12. anonymous

oh, so you get r^(n-1)+r^(n-2)+....r^3+r^2+r+1

13. anonymous

yeah

14. freckles

oops I didn't mean to write n-1 again

15. anonymous

yes I see, no need correction

16. freckles

$r^{n-1}+r^{n-2}+r^{n-\color{red}3}+ \cdots +r^{3}+r^2+r+1=\frac{r^n-1}{r-1}$

17. freckles

ok I did it anyways :p

18. 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

19. freckles

$S_n=a_1+ra_1+r^2a_1+r^3a_1 \cdots +r^{n-1}a_1 \\S_n= a_1(1+r+r^2+r^3+ \cdots r^{n-1}) \\ \\ S_n=a_1 \frac{r^n-1}{r-1} \\ \text{ or multiply both \top and bottom gives } S_n=a_1 \frac{1-r^n}{1-r}$

20. anonymous

the trick is to notice that if we let $$S_n$$ denote the sum of $$n$$ terms we get: $$S_n=a_1+a_1 r+a_1 r^2+\dots+a_1r^{n-1}\\r S_n=a_1 r+a_1 r^2+\dots+a_1r^{n-1}+a_1 r^n\\S_n-rS_n=a_1-a_1 r^n\\(1-r)S_n=a_1(1-r^n)\\S_n=a_1\frac{1-r^n}{1-r}$$

21. freckles

yes you are right so you can take that sum of the r things you wrote and multiply it by your initial

22. freckles

that's cute @oldrin.bataku

23. 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.

24. freckles

yep

25. 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.

26. anonymous

You really saved me. "Filled in the gap"

27. freckles

np didn't do much :)

28. 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