Question

The numerically greatest term in the expansion of (3x - 16y)¹⁵ when x = ²⁄₃ and y = ³⁄₂ is ?

• A. 13
• B. 14
• C. 15
• D. 16

Hint:

For largest term in \((a+b)^n\), one often uses the ratio of consecutive terms method: compare \(\frac{T_{r+1}}{T_r}\).
- After substituting numeric values for \(x,y\), check which \(r\) maximizes the magnitude.

Answer 1

The Correct Option is C

Step 1: General term.
In \((3x - 16y)^{15}\), the \((r+1)\)th term (using binomial expansion) is \[ T_{r+1} = \binom{15}{r} (3x)^{15-r} \,(-16y)^r. \] We want to find \(r\) that maximizes \(\bigl|T_{r+1}\bigr|\) for \(x=\frac{2}{3}\) and \(y=\frac{3}{2}\). Step 2: Substitute \(x\) and \(y\).
\[ T_{r+1} = \binom{15}{r} \,3^{15-r}\,\left(\frac{2}{3}\right)^{15-r} \,\bigl(-16\bigr)^r \,\left(\frac{3}{2}\right)^r. \] Simplify constants to get an expression in \(r\). 

Step 3: Ratio test for consecutive terms.
Compare \(\bigl|T_{r+1}\bigr|\) and \(\bigl|T_{r+2}\bigr|\). The ratio test helps pinpoint where the maximum occurs. Typically we check integer \(r\)-values near the ratio \(\approx 1\). 

Step 4: Conclusion.
Detailed algebra shows the maximum absolute term occurs for \(r=14\). That corresponds to the \((r+1)=(14+1)\) = 15\textsuperscript{th} term. Thus