JEE Main 2022 — Binomial Theorem Question with Solution

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

If T₉ is numerically greatest term

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

$${}^n{C}_7{3}^{n-7}(6x{)}^7\le {}^n{C}_8{3}^{n-8}(6x{)}^8\ge {}^n{C}_9{3}^{n-9}(6x{)}^9$$

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

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

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

$$\frac{29}{3}\le n$$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

$$r=3\Rightarrow {}^{10}{C}_3{3}^7.6^3$$

$$\therefore$$ $$k=\frac{{}^{10}{C}_6}{{}^{10}{C}_3}.\frac{3^4{.6}^6}{3^7{.6}^3}=\frac{10!7!3!}{6!4!10!}.8$$

$$\Rightarrow k=14$$

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