Question

@amistre64

Ok so you used the Geometric sequence:
\[B_n=BK^n-P*\text{ Geoemtric sum }\]
\[B_n=BK^n-P*\frac{k^n-1}{(k-1}\]

Shouldnt it be:
\[B_n=BK^n-P*\frac{1-k^n}{1-k}\]

Answer 1

lol, doesnt matter, you are confusing form with mathical logic

\[\frac{k^n-1}{k-1}\implies~\frac{-1}{-1}\frac{k^n-1}{k-1}\implies~\frac{1-k^n}{1-k}\]

Answer 2

ohh ok gotcha

Answer 3

Thanks

Answer 4

its just easier in terms of k in the denominator

1 + r - 1 = r

Answer 5

ya that i saw

Answer 6

K thanks :)

Answer 7

yep

Answer 8

just for fun ... spose P changes by some rule; \(P_{n+1} = P_{n}(1+r)\), we increase the payment by r% each time.

B0 = B

B1 = Bk - P

B2 = Bk^2 - Pk -P(1+r)

B3 = Bk^3 - Pk^2 -P(1+r)k - P(1+r)^2

let 1+r = m, and lets rewrite as


B0 = B

B1 = Bk - Pk^0m^0

B2 = Bk^2 - Pk^1m^0 - Pk^0m^1

B3 = Bk^3 - Pk^2m^0 - Pk^1m^1 - Pk^0m^2

we do have a pattern can we make a rule for the summation?

\[S = k^0m^{n-1} + k^1m^{n-2} + k^2m^{n-3}+...+ k^{n-2}m^1 + k^{n-1}m^0\]
\[-km^{-1}S = - k^1m^{n-2} - k^2m^{n-3}-...- k^{n-2}m^1 - k^{n-1}m^0 - k^nm^{-1}\]

\[S = \frac{m^{n-1} - k^nm^{-1}}{1-km^{-1}}\]

\[B_n = Bk^n - P\frac{m^{n-1} - k^nm^{-1}}{1-km^{-1}}\]
\[B_n = Bk^n - P\frac{m^{n} - k^n}{m-k}\]

when r=0, m=1

if r=.1/12, then what is our inital basis for a payment? :)

Answer 9

if the payments are truly variable, can they be determined before hand? or do we setup a variable payment schedule like 10, 12, 15, 20, 21, 32, 33, 36 ....

if we setup a payment schedule it may be best to just spreadsheet it

Answer 10

Well if its variable then i dont think its possible to find a general form
You are stuck using the following formula :
\[PV=\frac{P_1}{1+r}+\frac{P_2}{(1+r)^2}+...+ \frac{P_n}{(1+r)^n}\]

Answer 11

Anddd love what you did above ;)
I guess thats we can consider that the "variation" with payments :P