Question

$5000 is deposited into an account paying 9% interest compounded monthly. Find the difference equation showing how to compute yn from yn-1 and the initial value

Answer 1

I believe that we would say: \[
y_n = y_{n-1}(1+0.09)
\]

Answer 2

Actually, make that: \[
y_n = y_{n-1}(1+0.09) + 5000
\]

Answer 3

Likewise, we could say: \[
y_{n-1} = y_{n-2}(1+0.09) + 5000
\]So if we substitute this in we would get: \[
y_n=[y_{n-2}(1+0.09)+5000](1+0.09)+5000
\]

Answer 4

Doing this \(n-1\) times, we would get: \[
y_n = y_{n-n}(1.09)^n+\sum_{k=0}^{n-1}(1.09)^k5000
\]

Answer 5

Remember that \(y_{n-n} = y_0\) which is the initial amount.

Answer 6

Also :\[
\sum_{k=0}^{n-1}(1.09)^k5000 = 5000\sum_{k=0}^{n-1}(1.09)^k
\]Which is a simple geometric series.\[
5000\sum_{k=0}^{n-1}(1.09)^k = 5000\left[\frac{(1.09)^n-1}{1.09-1}\right]
\]