Let for the $9^{\text {th }}$ term in the binomial expansion of $(3+6 x)^{\mathrm{n}}$, in the increasing powers of $6 x$, to be the greatest for $x=\frac{3}{2}$, the least value of $\mathrm{n}$ is $\mathrm{n}_{0}$. If $\mathrm{k}$ is the ratio of the coefficient of $x^{6}$ to the coefficient of $x^{3}$, then $\mathrm{k}+\mathrm{n}_{0}$ is equal to :

<p>${(3 + 6x)^n} = {3^n}{(1 + 2x)^n}$</p> <p>If T₉ is numerically greatest term</p> <p>$\therefore$ ${T_8} \le {T_9} \le {T_{10}}$</p> <p>$${}^n{C_7}{3^{n - 7}}{(6x)^7} \le {}^n{C_8}{3^{n - 8}}{(6x)^8} \ge {}^n{C_9}{3^{n - 9}}{(6x)^9}$$</p> <p>$$ \Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {(n - 8)!8!}}3\,.\,(6x) \ge {{n!} \over {(n - 9)!9!}}{(6x)^2}$$</p> <p>$$ \Rightarrow \underbrace {{9 \over {(n - 7)(n - 8)}}}_{} \le \underbrace {{{18\left( {{3 \over 2}} \right)} \over {(n - 8)8}} \ge {{36} \over {9.8}}{9 \over 4}}_{}$$</p> <p>$72 \le 27(n - 7)$ and $27 \ge 9(n - 8)$</p> <p>${{29} \over 3} \le n$and $n \le 11$</p> <p>$\therefore$ ${n_0} = 10$</p> <p>For ${(3 + 6x)^{10}}$</p> <p>${T_{r + 1}} = {}^{10}{C_r}$</p> <p>${3^{10 - r}}{(6x)^r}$</p> <p>For coeff. of x⁶</p> <p>$r = 6 \Rightarrow {}^{10}{C_6}{3^4}{.6^6}$</p> <p>For coeff. of x³</p> <p>$r = 3 \Rightarrow {}^{10}{C_3}{3^7}{.6^3}$</p> <p>$\therefore$ $$k = {{{}^{10}{C_6}} \over {{}^{10}{C_3}}}.{{{3^4}{{.6}^6}} \over {{3^7}{{.6}^3}}} = {{10!7!3!} \over {6!4!10!}}.8$$</p> <p>$\Rightarrow k = 14$</p> <p>$\therefore$ $k + {n_0} = 24$</p>