Question

The numerically greatest term in the expansion of $ (x + 3y)^{13} $, when $ x = \frac{1}{2},\ y = \frac{1}{3} $, is

• A. \( \binom{13}{9} \left( \frac{1}{3} \right)^4 \)
• B. \( \binom{13}{4} \left( \frac{1}{2} \right)^9 \)
• C. \( \binom{13}{9} \left( \frac{1}{2} \right)^4 \)
• D. \( \binom{13}{10} \left( \frac{1}{2^4} \right) \)

Hint:

To find numerically greatest term, maximize \( T_r \) using ratio test or approximate \( \frac{r}{n+1} \approx \frac{ax}{ay} \).

Answer 1

The Correct Option is C

General term: \[ T_{r+1} = \binom{13}{r} (x)^{13 - r} (3y)^r \Rightarrow T_{r+1} = \binom{13}{r} \left( \frac{1}{2} \right)^{13 - r} \left( 1 \right)^r = \binom{13}{r} \left( \frac{1}{2} \right)^{13 - r} \] To maximize this term numerically, we use: \[ \frac{T_{r+1}}{T_r}>1 \Rightarrow \text{solve for maximum term} \Rightarrow \text{Using derivative logic or estimate, max occurs at } r = 9 \Rightarrow T_{10} = \binom{13}{9} \left( \frac{1}{2} \right)^4 \]