Key Concepts and Formulas

• Sum of a Finite Geometric Progression (GP): The sum of the first $n$ terms of a GP is given by $S_n = \frac{a(1 - r^n)}{1 - r}$, where $a$ is the first term and $r$ is the common ratio ($r \neq 1$).
• Inequality Manipulation: When multiplying or dividing an inequality by a negative number, the direction of the inequality sign must be reversed.
• Properties of Exponents: For a negative base $(-x)$, $(-x)^n$ is positive if $n$ is even and negative if $n$ is odd.

---

Step-by-Step Solution

Step 1: Express $A_n$ in closed form.

The given series for $A_n$ is: $A_n = \left(\frac{3}{4}\right) - \left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^3 - \dots + (-1)^{n-1} \left(\frac{3}{4}\right)^n$ This is a finite geometric progression. The first term is $a = \frac{3}{4}$. The common ratio is $r = \frac{-\left(\frac{3}{4}\right)^2}{\frac{3}{4}} = -\frac{3}{4}$. Using the GP sum formula $S_n = \frac{a(1 - r^n)}{1 - r}$: $A_n = \frac{\frac{3}{4}\left(1 - \left(-\frac{3}{4}\right)^n\right)}{1 - \left(-\frac{3}{4}\right)} = \frac{\frac{3}{4}\left(1 - \left(-\frac{3}{4}\right)^n\right)}{1 + \frac{3}{4}} = \frac{\frac{3}{4}\left(1 - \left(-\frac{3}{4}\right)^n\right)}{\frac{7}{4}}$ Simplifying, we get: $A_n = \frac{3}{7}\left(1 - \left(-\frac{3}{4}\right)^n\right)$

Step 2: Formulate and simplify the inequality $B_n > A_n$.

We are given $B_n = 1 - A_n$. The inequality $B_n > A_n$ becomes: $1 - A_n > A_n$ $1 > 2A_n$ $A_n < \frac{1}{2}$ Substitute the closed form of $A_n$: $\frac{3}{7}\left(1 - \left(-\frac{3}{4}\right)^n\right) < \frac{1}{2}$ Multiply by $\frac{7}{3}$: $1 - \left(-\frac{3}{4}\right)^n < \frac{7}{6}$ Subtract 1 from both sides: $-\left(-\frac{3}{4}\right)^n < \frac{7}{6} - 1$ $-\left(-\frac{3}{4}\right)^n < \frac{1}{6}$ Multiply by $-1$ and reverse the inequality sign: $\left(-\frac{3}{4}\right)^n > -\frac{1}{6}$

Step 3: Analyze the inequality based on the parity of $n$.

We need to find the least odd natural number $p$ such that $\left(-\frac{3}{4}\right)^n > -\frac{1}{6}$ for all $n \ge p$.

• Case 1: $n$ is even. If $n$ is even, $\left(-\frac{3}{4}\right)^n = \left(\frac{3}{4}\right)^n$. Since $\left(\frac{3}{4}\right)^n$ is always positive, and $-\frac{1}{6}$ is negative, the inequality $\left(\frac{3}{4}\right)^n > -\frac{1}{6}$ is true for all even natural numbers $n$.

• Case 2: $n$ is odd. If $n$ is odd, $\left(-\frac{3}{4}\right)^n = -\left(\frac{3}{4}\right)^n$. The inequality becomes: $-\left(\frac{3}{4}\right)^n > -\frac{1}{6}$ Multiply by $-1$ and reverse the inequality sign: $\left(\frac{3}{4}\right)^n < \frac{1}{6}$ We need to find the smallest odd integer $n$ for which this inequality holds. Let's test values:

  • $n=1$: $\left(\frac{3}{4}\right)^1 = 0.75$. $0.75 \not< \frac{1}{6}$.
  • $n=3$: $\left(\frac{3}{4}\right)^3 = \frac{27}{64} \approx 0.42$. $0.42 \not< \frac{1}{6}$.
  • $n=5$: $\left(\frac{3}{4}\right)^5 = \frac{243}{1024} \approx 0.237$. $0.237 \not< \frac{1}{6}$.
  • $n=7$: $\left(\frac{3}{4}\right)^7 = \frac{2187}{16384} \approx 0.133$. $0.133 \not< \frac{1}{6}$.
  • $n=9$: $\left(\frac{3}{4}\right)^9 = \frac{19683}{262144} \approx 0.075$. $0.075 < \frac{1}{6}$. Since $\left(\frac{3}{4}\right)^n$ decreases as $n$ increases, the inequality $\left(\frac{3}{4}\right)^n < \frac{1}{6}$ holds for all odd $n \ge 9$.

Step 4: Determine the least odd natural number $p$.

The condition $B_n > A_n$ holds for all even $n$ and for all odd $n \ge 9$. We need to find the least odd natural number $p$ such that for all $n \ge p$, the inequality $B_n > A_n$ holds. This means that $p$ must be an odd number, and all integers from $p$ upwards must satisfy the condition. If $p=7$, then for $n=7$ (which is $\ge p$), the inequality $\left(\frac{3}{4}\right)^7 < \frac{1}{6}$ does not hold. So $p$ cannot be 7. If $p=9$, then for all $n \ge 9$:

• If $n$ is even, $B_n > A_n$ holds.
• If $n$ is odd, since $n \ge 9$, $B_n > A_n$ holds. Therefore, for $p=9$, the condition $B_n > A_n$ holds for all $n \ge 9$. Since 9 is the smallest odd number for which this is true, $p=9$.

---

Common Mistakes & Tips

• Sign Errors: Be extremely careful when multiplying or dividing inequalities by negative numbers. Always reverse the inequality sign.
• Parity Analysis: The behavior of terms with negative bases raised to powers depends crucially on whether the exponent is even or odd. Treat these cases separately.
• Testing Values: When dealing with inequalities involving powers, testing small integer values is an effective way to find the threshold at which the inequality begins to hold.

---

Summary

The problem requires finding the smallest odd natural number $p$ such that for all $n \ge p$, $B_n > A_n$. We first found a closed-form expression for $A_n$ as a geometric progression. Then, we converted the condition $B_n > A_n$ into an inequality involving $n$. By analyzing the inequality separately for even and odd values of $n$, we determined that the condition holds for all even $n$ and for odd $n \ge 9$. The least odd natural number $p$ that guarantees the condition for all $n \ge p$ is therefore 9.

The final answer is $\boxed{9}$.