Answer
See explanation
Work Step by Step
We wish to prove by mathematical induction that for all integers \( k \ge 1 \) the sequence defined by
\[
g_1 = 1 \quad\text{and}\quad g_k = \frac{g_{k-1}}{g_{k-1} + 2}\quad\text{for } k \ge 2,
\]
has the closed-form solution
\[
\boxed{g_k = \frac{1}{2^k - 1}}.
\]
---
### **Base Case** \((k = 1)\)
For \( k = 1 \), the formula gives
\[
g_1 = \frac{1}{2^1 - 1} = \frac{1}{2-1} = 1.
\]
This matches the given initial condition \(g_1 = 1\).
---
### **Inductive Step**
**Inductive Hypothesis:**
Assume that for some integer \( k \ge 2 \) the formula holds for \( k-1 \); that is,
\[
g_{k-1} = \frac{1}{2^{\,k-1} - 1}.
\]
**Goal:**
We must show that under this assumption,
\[
g_k = \frac{1}{2^k - 1}.
\]
**Proof:**
By the recurrence relation, we have
\[
g_k = \frac{g_{k-1}}{g_{k-1} + 2}.
\]
Substitute the inductive hypothesis into the formula:
\[
g_k = \frac{\displaystyle \frac{1}{2^{k-1} - 1}}{\displaystyle \frac{1}{2^{k-1} - 1} + 2}.
\]
To combine the terms in the denominator, write \(2\) with a common denominator:
\[
\frac{1}{2^{k-1} - 1} + 2 = \frac{1}{2^{k-1} - 1} + \frac{2(2^{k-1} - 1)}{2^{k-1} - 1}
= \frac{1 + 2(2^{k-1} - 1)}{2^{k-1} - 1}.
\]
Simplify the numerator:
\[
1 + 2(2^{k-1} - 1) = 1 + 2 \cdot 2^{k-1} - 2 = 2 \cdot 2^{k-1} - 1 = 2^k - 1.
\]
Thus, the denominator becomes
\[
\frac{2^k - 1}{2^{k-1} - 1}.
\]
Now, we substitute back into the expression for \( g_k \):
\[
g_k = \frac{\displaystyle \frac{1}{2^{k-1} - 1}}{\displaystyle \frac{2^k - 1}{2^{k-1} - 1}}
= \frac{1}{2^k - 1}.
\]
This is exactly the formula we needed to prove for \( g_k \).
