Answer
See explanation
Work Step by Step
**Problem Statement (Rephrased)**
Show that for every integer \(n \ge 1\),
\[
\frac{1 + 3 + 5 + \cdots + (2n - 1)}{(2n + 1) + (2n + 3) + \cdots + (4n - 1)} \;=\; \frac{1}{3}.
\]
---
## 1. Recognize the Sums
1. **Numerator**: \(1 + 3 + 5 + \cdots + (2n - 1)\)
This is the sum of the first \(n\) odd numbers. A well-known result (which can be proved by induction or by formula) is:
\[
1 + 3 + 5 + \cdots + (2n - 1) \;=\; n^2.
\]
So the numerator is \(n^2\).
2. **Denominator**: \((2n+1) + (2n+3) + \cdots + (4n - 1)\)
These are \(n\) consecutive odd numbers starting at \(2n+1\) and ending at \(4n - 1\). One quick way to sum \(n\) equally spaced numbers is to multiply the average of the first and last terms by the number of terms.
- **First term**: \(2n + 1\)
- **Last term**: \(4n - 1\)
- **Number of terms**: \(n\) (because we step by 2 from \(2n+1\) to \(4n-1\))
- **Average** of the first and last terms:
\[
\frac{(2n+1) + (4n - 1)}{2} \;=\; \frac{6n}{2} \;=\; 3n.
\]
- Hence, the sum of these \(n\) odd numbers is
\[
n \times \bigl(\text{average}\bigr) \;=\; n \times (3n) \;=\; 3n^2.
\]
So the denominator is \(3n^2\).
---
## 2. Form the Fraction
Putting these results together:
\[
\frac{1 + 3 + 5 + \cdots + (2n - 1)}{(2n + 1) + (2n + 3) + \cdots + (4n - 1)}
\;=\;
\frac{n^2}{3n^2}
\;=\;
\frac{1}{3}.
\]
Thus, for all integers \(n \ge 1\), the given fraction equals \(\tfrac13\). This completes the proof.
---
### Alternative Approach
**Induction on \(n\)**:
- Base Case \((n=1)\):
\(\tfrac{1}{3} = \tfrac{1}{3}\) is trivially true.
- Inductive Step: Show that going from \(n\) to \(n+1\) preserves the fraction’s value of \(\tfrac13\).
We have to prove that
$3[1+3+...+(2n-1)]=(2n+1)+...+(4n-1)$
Consider this is true for $k$ and prove for $k+1$.
