Question

Let for the 9th term in the binomial expansion of (3 + 6x)n, in the increasing powers of 6 x , to be the greatest for x = 3/2, the least value of n is n 0. If k is the ratio of the coefficient of x 6 to the coefficient of x 3, then k + n0 is equal to :

Answer 1

$\begin{array}{l}\left(3 + 6x\right)^n = 3^n\left(1 + 2x\right)^n\end{array}$
If T 9 is numerically greatest term
$\begin{array}{l}\therefore T_8 \le T_9 \geq T_{10}\end{array}$
$\begin{array}{l}nC_7 3^{n–7} \left(6x\right)^7\le nC_8 3^{n–8} \left(6x\right)^8 \geq nC_9 3^{n–9} \left(6x\right)^9\end{array}$
$\begin{array}{l} \Rightarrow\ \frac{n!}{\left(n-7\right)!7!}9\leq \frac{n!}{\left(n-8\right)!8!}3\cdot\left(6x\right)\geq\frac{n!}{\left(n-9\right)!9!}\left(6x\right)^2\end{array}$
$\begin{array}{l} \Rightarrow\ \underbrace{\frac{9}{\left(n-7\right)\left(n-8\right)}}\leq\underbrace{\frac{18\left(\frac{3}{2}\right)}{\left(n-8\right)8}}\geq \frac{36}{9.8}\frac{9}{4}\end{array}$
$\begin{array}{l}72 \le 27\left(n – 7\right) ~\text{and}~ 27 \geq 9\left(n – 8\right)\end{array}$
$\begin{array}{l} \frac{29}{3}\leq n\ \text{and}\ n \le 11\end{array}$
$\begin{array}{l}\therefore n_0 = 10\end{array}$
$\begin{array}{l}\text{For} \left(3 + 6x\right)^{10}\end{array}$
$\begin{array}{l}T_{r + 1} = ^{10}C_r ~~~~~~~~~~~~3^{10 – r} \left(6x\right)^r\end{array}$
$\begin{array}{l}\text{For coeff. of }x^6\\\ r = 6 \Rightarrow ^{10}C_63^4.6^6\end{array}$
$\begin{array}{l}\text{For coeff. of} x^3\\\ r = 3 \Rightarrow ^{10}C_33^7.6^3\end{array}$
$\begin{array}{l} \therefore\ k=\frac{^{10}C_6}{^{10}C_3}\cdot\frac{3^4\cdot 6^6}{3^7\cdot 6^3}=\frac{10!7!3!}{6!4!10!}\cdot 8 \end{array}$
$\begin{array}{l}\Rightarrow k = 14\\\ \therefore k + n_0 = 24\end{array}$