If the numerically greatest term in the expansion of $(2-3 \mathrm{x})^9$ when $\mathrm{x}=1$ is…

If the numerically greatest term in the expansion of $(2-3 \mathrm{x})^9$ when $\mathrm{x}=1$ is $\mathrm{P}_1^\alpha \mathrm{P}_2^\beta \mathrm{P}_3^\gamma \mathrm{P}_4^\delta\left(\mathrm{P}_1 < \mathrm{P}_2 < \mathrm{P}_3 < \mathrm{P}_4\right.$ are the first four prime numbers), then $\alpha+\beta+\gamma+\delta=$
  1. $13$
  2. $12$
  3. $14$
  4. $11$

Solution

For greatest term in expansion: $\begin{aligned} & \mathrm{T}_{r+1}>\mathrm{T}_r \\ & \Rightarrow{ }^9 C_r 2^r(|-3 x|)^{9-r}>{ }^9 C_{r-1} 2^{r-1}(|-3 x|)^{9-(r-1)} \\ & \Rightarrow \frac{9 !}{r !(9-r) !} 2 \geq \frac{9 !}{(r-1) !(9-r+1) !} 3 x \\ & \Rightarrow \frac{2}{r} \geq \frac{3}{9-r+1}\{\because x=1\} \\ & \Rightarrow 20-2 r \geq 3 r \Rightarrow r \geq 4 \\ & \therefore r=4 \\ & T_5={ }^9 C_4(2)^4[-3]^5=\frac{9 !}{4 ! \times 5 !}(2)^4 \cdot 3^5 \\ & =\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 6} \times 2^4 \times 3^5=3^7 \cdot 2^5 \cdot 7 \\ & \therefore \alpha=7, \beta=5, \gamma=1, \delta=0 \\ & \Rightarrow \alpha+\beta+\gamma+\delta=7+5+1+0=13 \end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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