Question

If \( p = \frac{1}{3} \) and \( q = \frac{2}{3} \), then what is the smallest odd \( n \) such that \( a_n + b_n<0.01 \)?

• A. 7
• B. 13
• C. 11
• D. 9
• E. 15

Hint:

Always check borderline values when dealing with inequalities in sequences, especially when the exponent pattern involves odd/even restrictions.

Answer 1

The Correct Option is C

From recurrence, for odd \( n \):
\( a_n + b_n = p(pq)^{\frac{n-1}{2} - 1}(p+q) \).
Given \( p = \frac{1}{3}, q = \frac{2}{3} \), \( pq = \frac{2}{9} \), \( p+q = 1 \).
Thus: \[ a_n + b_n = \frac{1}{3} \left( \frac{2}{9} \right)^{\frac{n-3}{2}}. \] We require \( a_n + b_n<0.01 \): \[ \frac{1}{3} \left( \frac{2}{9} \right)^{\frac{n-3}{2}}<0.01. \] Multiply by 3: \[ \left( \frac{2}{9} \right)^{\frac{n-3}{2}}<0.03. \] Take logs: \[ \frac{n-3}{2} \log\left( \frac{2}{9} \right)<\log(0.03). \] Since log is negative, inequality reverses: \[ \frac{n-3}{2}>\frac{\log(0.03)}{\log(2/9)}. \] Numerically: \(\log_{10}(0.03) \approx -1.52288\), \(\log_{10}(2/9) \approx -0.65321\).
Ratio ≈ 2.33 → \( \frac{n-3}{2}>2.33 \) → \( n - 3>4.66 \) → \( n>7.66 \).
Since \( n \) is odd, smallest odd \( n = 9 \) → but check actual value:
For \( n=9 \): \( \frac{1}{3} (2/9)^3 \approx 0.01096>0.01 \), fails.
Next odd \( n = 11 \): \( \frac{1}{3} (2/9)^4 \approx 0.00243<0.01 \), works.