series and sequences question

i've gotta be honest, i dislike how they did part a but their method will help you with solving part b.

for part a, i (and i think you too) did:
( ( 150,000 * 1.06 - 10,000 ) * 1.06 - 10,000 ) * 1.06 - 10,000
= ( 150,000 * 1.06 - 10,000 ) * (1.06)^2 - 10,000 * 1.06 - 10,000
= 150,000 * (1.06)^3 - 10,000 * (1.06)^2 - 10,000 * 1.06 - 10,000
which feels like the easiest solution to me, but it doesn't help with part b.


the mark scheme has instead represented the total mortgage as the sum of a geometric series and a recurrence relation, by doing:
[the total mortgage] = [the original amount owed, increasing due to interest each year] - [the amount payed off, which increases by 10,000 and increases due to interest each year]
we'll call the original amount series A, and the payed off amount relation B.

A at the end of year 0 = 150,000
A ... 1 = 150,000 * 1.06
A ... 2 = 150,000 * (1.06)^2
therefore, A at the end of year n = 150,000 * (1.06)^n

B at the end of year 0 = 0
B ... 1 = [B at the end of year 0] * 1.06 + 10,000
B ... 2 = [B ... 1] * 1.06 + 10,000
therefore, B at the end of year n = [B at the end of year (n-1)] * 1.06 + 10,000
and B ... (n+1) = [B ... n] * 1.06 + 10,000

you can substitute in n=3 and multiply out B to do part a.

then to do part b, i can't be 100% sure on what form the markscheme wants an answer in but i'm guessing it won't be happy with just:
150,000 * (1.06)^n - [B at the end of year (n-1)] * 1.06 + 10,000

so, we need to get the recurrence relation ([B ... n] in terms of [B ... (n-1)]) into a 'closed form solution' ([B ... n] in terms of [n]).
i'm a little iffy on this topic, but if i'm understanding my textbook correctly then to solve this you need to do something similar to solving a second order DE.

(pls pretend for the B(n+1) and B(n) the n+1 and n are in subscript)

for a first order recurrence relation B(n+1) = 1.06 * u(n) + 10,000:

step 1: find the complementary function c(n):
a - start with the reduced equation B(n+1) = 1.06 * u(n) and replace B(n) with r^n (where r is a non-zero constant)
b - this gives the auxiliary equation r^(n+1) = 1.06 * r^n, therefore r = 1.06
c - the complementary function is then c(n) = A * 1.06^n (where A is a constant (unrelated to the A we had earlier))

step 2: find the particular solution p(n):
a - replace the [+ 10,000] with a general function of the same form (as + 10,000 is a constant our p(n) = a (where a is a constant))
b - substitute B(n) = p(n) to give: a = 1.06 * a + 10,000 (p(n) is a function of n, but as it contains no n's p(n) = p(n+1) = etc)
c - therefore a = -10,000 / (1.06 - 1) = - 500,000 / 3

step 3: find the general (closed form) solution B(n):
a - B(n) = c(n) + p(n) = A * 1.06^n - (500,000 / 3)
b - apply the initial conditions (in our case it's that B(0) = 0 and B(1) = 10,000) to find A:
c - therefore 0 = A * 1.06^0 - (500,000 / 3) therefore (500,000 / 3) = A * 1 therefore A = (500,000 / 3)
d - therefore B(n) = (500,000 / 3) * 1.06^n - (500,000 / 3)

so i think that should the solution to part b should be A(n) - B(n) = 150,000 * 1.06^n - ( (500,000 / 3) * 1.06^n - (500,000 / 3) )
although the markscheme might want a more simplified form

for part c, you want to do A(n) - B(n) = 0
therefore 0 = 150,000 * 1.06^n - ( (500,000 / 3) * 1.06^n - (500,000 / 3) )
= ( 150,000 - (500,000 / 3) ) * 1.06^n + (500,000 / 3)
therefore (50,000 / 3) * 1.06^n = (500,000 / 3)
therefore 1.06^n = 10 therefore n = log1.06 (10) = 39.516...
therefore (we have to round up) it will take 40 years to pay off the mortgage

it's almost 2am so i don't have the energy to check my working, but even if i misplaced a 0 somewhere this should at least show you how to do this question. i hope this helped!

I am not sure what 2nd order DE and auxiliary equations are.. I don't think it is part of my maths spec but thanks for your help!!