Let for the 9^th term in the binomial expansion of (3 + 6x)ⁿ, in the increasing powers of 6x, to be the greatest for $\mathrm{x}=\frac{3}{2}$, the least value of n is n₀. If k is the ratio of the coefficient of x⁶ to the coefficient of x³, then k + n₀ is equal to:

$\begin{array}{l}(3+6\mathrm{x}{)}^{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0{3}^{\mathrm{n}}{+}^{\mathrm{n}}{\mathrm{C}}_1{3}^{\mathrm{n}-1}(6\mathrm{x}{)}^1+\dots \\ {{\mathrm{T}}_{\mathrm{r}+1}}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{3}^{\mathrm{n}-\mathrm{r}}\cdot \left(6\mathrm{x}\right)\mathrm{r}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{3}^{\mathrm{n}-\mathrm{r}}\cdot 6^{\mathrm{r}}\cdot {\mathrm{x}}^{\mathrm{r}}\\ {=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{3}^{\mathrm{n}-\mathrm{r}}\cdot 3^{\mathrm{r}}\cdot 2^{\mathrm{r}}\cdot {\left(\frac{3}{2}\right)}^{\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{3}^{\mathrm{n}}\cdot 3^{\mathrm{r}}\left[\text{ for }\mathrm{x}=\frac{3}{2}\right]\end{array}$

T₉ is greatest of $\mathrm{x}=\frac{3}{2}$

So, T₉ > T₁₀ and T₉ > T₈

(concept of numerically greatest term)

Here, $\frac{{\mathrm{T}}_9}{{\mathrm{T}}_{10}}>1\text{ and }\frac{{\mathrm{T}}_9}{{\mathrm{T}}_8}>1$

$\Rightarrow \frac{{ }^{\mathrm{n}}{\mathrm{C}}_8{3}^{\mathrm{n}}\cdot 3^8}{{ }^{\mathrm{n}}{\mathrm{C}}_9{3}^{\mathrm{n}}\cdot 3^9}>1\text{ and }\frac{{ }^{\mathrm{n}}{\mathrm{C}}_8{3}^{\mathrm{n}}\cdot 3^8}{{ }^{\mathrm{n}}{\mathrm{C}}_7{3}^{\mathrm{n}}\cdot 3^7}>1$

and $\frac{{ }^{\mathrm{n}}{\mathrm{C}}_8}{{ }^{\mathrm{n}}{\mathrm{C}}_7}>\frac{1}{3}$

and $\frac{\mathrm{n}-7}{8}>\frac{1}{3}$

$\Rightarrow \frac{29}{3}<\mathrm{n}<11\Rightarrow \mathrm{n}=10={\mathrm{n}}_0$

So, $\text{ in }(3+6\mathrm{x}{)}^{\mathrm{n}}\text{ for }\mathrm{n}={\mathrm{n}}_0=10$

i.e., $\text{ in }(3+6\mathrm{x}{)}^{10},\text{ here }{\mathrm{T}}_{\mathrm{r}+1}{=}^{10}{\mathrm{C}}_{\mathrm{r}}{3}^{10-\mathrm{r}}{6}^{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}$

$\begin{array}{l}{\mathrm{T}}_7{=}^{10}{\mathrm{C}}_6{3}^4\cdot 6^6\cdot {\mathrm{x}}^6=210\cdot 3^{10}\cdot 2^6{\mathrm{x}}^6\\ {\mathrm{T}}_4{=}^{10}{\mathrm{C}}_3{3}^7{6}^3{\mathrm{x}}^3=120\cdot 3^{10}\cdot 2^3{\mathrm{x}}^3\end{array}$

Ratio of coefficient of x⁶ and coefficient of x³ = k

$\therefore \mathrm{k}=\frac{210\cdot 3^{10}{2}^6}{120{.3}^{10}\cdot 2^3}=\frac{7}{4}\times 2^3=14$

So, $\mathrm{k}+{\mathrm{n}}_0=14+10=24$