Question

please help!!!!!
Brian has a 25-year adjustable rate mortgage that has a rate of 3.2% for the first 4 years. The amount of the mortgage is $179,000. What is the monthly payment during this initial period?

$867.58

$596.67

$1,017.89

$615.76

Answer 1

the intitial rate would be $615.76 per month

Answer 2

P = Bk^(25*12) (1-k)/(1-k^(25*12)), k=1+.032/12

Answer 3

http://www.wolframalpha.com/input/?i=179000k%5E%2825*12%29+%281-k%29%2F%281-k%5E%2825*12%29%29%2C+k%3D1%2B.032%2F12

Answer 4

my mind has just been blown i did it wrong, and i know why thank you @amistre64

Answer 5

yep ....

Answer 6

Wait so its not D ?

Answer 7

why would it be D?

Answer 8

take my medal away and give it to @amistre64

Answer 9

I see it now, thank you so much!

Answer 10

how did you know that formula @amistre64 ಠ_ಠ

Answer 11

lol; i created it

Answer 12

I posted another, can you guys help?

Answer 13

i wrote out the recurrsion and simplified it; if i simplify it any more i get the textbooks formula ... but i find the textbook less intuitive

Answer 14

Amistre64 should im still trying to figure out the formula

Answer 15

ive never met anyone like you before, you are genuis

Answer 16

let k be that compounding mess to clean this up; B is balance and P is payment
\[B_o=B_0\]
\[B_1=B_0k - P\]
\[B_2=B_0k^2 - Pk -P\]
\[B_3=B_0k^3 - Pk^2 -Pk -P\]
this is a geometric sum of k on P
\[B_n=B_ok^n-P\frac{1-k^n}{1-k}\]
the nth Balance needs to be zero for the loan to be paid in full
\[0=B_ok^n-P\frac{1-k^n}{1-k}\]
solve for P
\[B_ok^n=P\frac{1-k^n}{1-k}\]
\[B_ok^n\frac{1-k}{1-k^n}=P\]

Answer 17

@amistre64 is very smart and good at these problems!!

Answer 18

I posted another if you can check it out

Answer 19

i have my moments :) but im outta time for now, ill give a quick look at it tho

Answer 20

that formula actually makes sense...mother of godಠ_ಠ

Answer 21

lol :)

Answer 22

yall have fun ... i might see wolverine this weekend