QJEEM22Ma57a6: JEE Main 2022 Mathematics - Class 11 JEEM22Ma57a6 Solution: Let for the term in the binomial expansion of , in the…

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

If T₉ is numerically greatest term

$\therefore$ ${T_8} \le {T_9} \le {T_{10}}$

${}^n{C_7}{3^{n - 7}}{(6x)^7} \le {}^n{C_8}{3^{n - 8}}{(6x)^8} \ge {}^n{C_9}{3^{n - 9}}{(6x)^9}$

$\Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {(n - 8)!8!}}3\,.\,(6x) \ge {{n!} \over {(n - 9)!9!}}{(6x)^2}$

$\Rightarrow \underbrace {{9 \over {(n - 7)(n - 8)}}}_{} \le \underbrace {{{18\left( {{3 \over 2}} \right)} \over {(n - 8)8}} \ge {{36} \over {9.8}}{9 \over 4}}_{}$

$72 \le 27(n - 7)$ and $27 \ge 9(n - 8)$

${{29} \over 3} \le n$ and $n \le 11$

$\therefore$ ${n_0} = 10$

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

${T_{r + 1}} = {}^{10}{C_r}$

${3^{10 - r}}{(6x)^r}$

For coeff. of x⁶

$r = 6 \Rightarrow {}^{10}{C_6}{3^4}{.6^6}$

For coeff. of x³

$r = 3 \Rightarrow {}^{10}{C_3}{3^7}{.6^3}$

$\therefore$ $k = {{{}^{10}{C_6}} \over {{}^{10}{C_3}}}.{{{3^4}{{.6}^6}} \over {{3^7}{{.6}^3}}} = {{10!7!3!} \over {6!4!10!}}.8$

$\Rightarrow k = 14$

$\therefore$ $k + {n_0} = 24$