So this is something really really simple but for some reason I honestly cannot figure out why this is wrong.
I was deriving the equation of the summation of a geometric series to the nth term because I hate memorizing.
So this is what I did: Let
$S = a + ar + ar^2 + ar^3 + ... + ar^n $ where a is some constant
First I subrated $a*r^n$. Getting:
(A)
$S - ar^n = a + ar + ar^2 + ... + ar^{n-1}$
Then,
(B)
$S - a = ar + ar^2 +ar^3 + ... + ar^n $
Those seemed reasonable, so then I divided equation B by r
(B_1)
$(S - a) / r = (ar + ar^2 + ... + ar^n) / r $
So that,
$ (S / r) - (a / r) = a + ar + ar^2 + ar^3 + ... ar^{n-1}$
Via distribution of (1/r)
Now we are and equality between A and B_1. Therefore:
(C)
$S - ar^n = (S/r) - (a/r) $
Which also seems reasonable. Then move both S terms and ar^n:
(C_1)
$S - (S/r) = ar^n - (a/r) $
Then factor out S:
(C_2)
$S (1 - (1/r)) = a(r^n - (1/r)) $
Finally divide:
(C_3)
$S = a(r^n - (1/r)) / (1 - (1/r)) $
Then I multiplied the right side by (-r/-r):
(F)
$S = a(1 - r^(n+1)) / (1 - r) $
as our final equation.
Of course the correct equation is: (CORRECT)
$S = a(1 - r^n) / (1 - r) $
if you try $S_{mine }= S_{correct}$, you eventually show that 1 = 0. Which is incorrect.
I understand the usual derivation of the equation, but it's bothering me that I can't see why this is wrong. Any help is greatly appreciated! Thanks!
You did nothing wrong. It is actually true that$$a+ar+ar^2+\cdots+ar^n=a\frac{1-r^{n+1}}{1-r}.$$Congratulations!