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 x= 3 2, the least value of n is…

Given,

$${(3+6x)}^{n}={3}^{n}{(1+2x)}^{n}$$(3+6x)n=3n(1+2x)n

Now if ${T}_{9}$T9​ is numerically greatest term, then ${T}_{8}\leq {T}_{9}\geq {T}_{10}$T8​≤T9​≥T10​

So, $C7n{3}^{n-7}{(6x)}^{7}\leq C8n{3}^{n-8}{(6x)}^{8}\geq C9n{3}^{n-9}{(6x)}^{9}$C7n3n−7(6x)7≤C8n3n−8(6x)8≥C9n3n−9(6x)9

$$\Rightarrow \frac{n!}{(n-7)!7!}9\leq \frac{n!}{(n-8)!8!}3\cdot (6x)\geq \frac{n!}{(n-9)!9!}{(6x)}^{2}$$⇒(n−7)!7!n!​9≤(n−8)!8!n!​3⋅(6x)≥(n−9)!9!n!​(6x)2

$$\Rightarrow \underset{⏟}{\frac{9}{(n-7)(n-8)}}\leq \underset{⏟}{\frac{18(\frac{3}{2})}{(n-8)8}\geq \frac{36}{9.8}\frac{9}{4}}$$⇒⏟(n−7)(n−8)9​​≤⏟(n−8)818(23​)​≥9.836​49​​

$\Rightarrow 72\leq 27(n-7)$⇒72≤27(n−7) and $27\geq 9(n-8)$27≥9(n−8)

$\Rightarrow \frac{29}{3}\leq n$⇒329​≤n and $n\leq 11$n≤11

So, ${n}_{0}=10$n0​=10

For ${(3+6x)}^{10}$(3+6x)10

Now ${T}_{r+1}=Cr10{3}^{10-r}{(6x)}^{r}$Tr+1​=Cr10310−r(6x)r

For coefficient of ${x}^{6}$x6

$$r=6\Rightarrow C610{3}^{4}\cdot {6}^{6}$$r=6⇒C61034⋅66

For coefficient of ${x}^{3}$x3

$$r=3\Rightarrow C310{3}^{7}\cdot {6}^{3}$$r=3⇒C31037⋅63

So, $k=\frac{C610}{C310}\cdot \frac{{3}^{4}\cdot {6}^{6}}{{3}^{7}\cdot {6}^{3}}=\frac{10!7!3!}{6!4!10!}\cdot 8$k=C310C610​⋅37⋅6334⋅66​=6!4!10!10!7!3!​⋅8

$$\Rightarrow k=14$$⇒k=14

$$\therefore k+{n}_{0}=24$$∴k+n0​=24