Answer
**(a)**
\[
\boxed{
Y_n = (E + c) \left( \frac{1 - m^n}{1 - m} \right) + m^n Y_0, \quad \text{for } n \ge 1
}
\]
**(b)**
If \(0 < m < 1\), then
\[
\boxed{
\lim_{n \to \infty} Y_n = \frac{E + c}{1 - m}
}
\]
Work Step by Step
### Given:
- Income in period \(k\): \(Y_k\)
- Consumption: \(C_k = c + m Y_{k-1}\)
- Total income: \(Y_k = C_k + E = c + m Y_{k-1} + E\)
So the recurrence is:
\[
\boxed{Y_k = (E + c) + m Y_{k-1}}, \quad \text{for } k \ge 1
\]
with initial value \(Y_0\) given.
---
### **(a) Use iteration to derive a closed-form formula for \(Y_n\)**
We’ll unfold the recurrence:
\[
\begin{aligned}
Y_1 &= (E + c) + m Y_0 \\
Y_2 &= (E + c) + m Y_1 = (E + c) + m[(E + c) + m Y_0] = (E + c)(1 + m) + m^2 Y_0 \\
Y_3 &= (E + c) + m Y_2 = (E + c)(1 + m + m^2) + m^3 Y_0 \\
\vdots \\
Y_n &= (E + c)(1 + m + m^2 + \cdots + m^{n-1}) + m^n Y_0
\end{aligned}
\]
The sum \(1 + m + m^2 + \cdots + m^{n-1}\) is a geometric series:
\[
\sum_{k=0}^{n-1} m^k = \frac{1 - m^n}{1 - m}, \quad \text{if } m \ne 1
\]
Therefore:
\[
\boxed{
Y_n = (E + c)\left(\frac{1 - m^n}{1 - m}\right) + m^n Y_0
}
\]
Now rewrite it to match the format requested in the problem:
Let’s multiply numerator and denominator of the first term by \(m\):
\[
\frac{1 - m^n}{1 - m} = \frac{m - m^{n+1}}{m(1 - m)} = \left(\frac{m^n - 1}{m - 1}\right)
\]
So we get:
\[
\boxed{
Y_n = (E + c) \left(\frac{m^n - 1}{m - 1} \right) + m^n Y_0, \quad \text{for } n \ge 1
}
\]
### **(b) Take the limit as \(n \to \infty\)**
We are told to assume \(0 < m < 1\). That means:
- \(\lim_{n \to \infty} m^n = 0\)
Apply this to the closed-form formula:
\[
Y_n = (E + c) \left( \frac{1 - m^n}{1 - m} \right) + m^n Y_0
\]
Take the limit:
\[
\lim_{n \to \infty} Y_n
= (E + c)\left( \frac{1 - 0}{1 - m} \right) + 0
= \frac{E + c}{1 - m}
\]
