Question

Numerically greatest term in the expansion of \((2x-3y)^n\) for \(x=\frac{3}{2}, y=\frac{1}{3}, n=6\) is

• A. 1215
• B. 1458
• C. 1024
• D. 2187 \bigskip

Hint:

For greatest term in binomial expansion, use ratio of consecutive terms instead of full expansion.

Answer 1

The Correct Option is B

Concept: General term: \[ T_{r+1}=\binom{n}{r}(2x)^{n-r}(-3y)^r \] Substitute values and find maximum term using ratio test.

Step 1: Substitute values.
\[ 2x=3,\quad 3y=1 \] So expression becomes: \[ (3-1)^6 \] But to find numerical greatest term, we use term ratio.

Step 2: Ratio of consecutive terms.
\[ \frac{T_{r+1}}{T_r} = \frac{6-r}{r+1}\cdot\frac{1}{3} \] Set: \[ \frac{T_{r+1}}{T_r}=1 \] \[ \frac{6-r}{r+1}=\!3 \] \[ 6-r=3r+3 \] \[ r=\frac{3}{2} \] So maximum at \(r=1\) or \(2\). Testing gives max at \(r=1\).

Step 3: Compute term.
\[ T_2=\binom{6}{1}3^5(-1)=6\times243\times(-1) \] Magnitude: \[ 1458 \] \[ \boxed{1458} \] Hence correct option: \[ \boxed{(B)}. \]