Question

$\large{\colorbox{gold}{\color{red}{Answer~the~following~question!}}}$​

Answer 1

$\orange{\mathbb{BINOMIAL\:EXPANSION}} \\ \\ \tt We\:know\:that \\ \\ \tt The\:expansion\:of\:{(x+y)}^{n} \: is \\ \\ \tt \boxed{\bold{\underline{\green{\tt {(x+y)}^{n} = {}^{n}{C}_{0}{(x)}^{n}{(y)}^{0} + {}^{n}{C}_{1}{(x)}^{n-1}{(y)}^{1} + ... }}}} \\ \\ \tt Also\:the\:(r+1)th\:term\:is\:given\:by \\ \\ \tt \boxed{\bold{\underline{\green{\tt {T}_{r+1} = {}^{n}{C}_{r}{(x)}^{n-r}{(y)}^{r} }}}} \\ \\ \tt So\:the\:{5}^{th} \:term\:in\: expansion\:of \: {(\frac{x}{3} - 3y)}^{7} \\ \\ \tt \implies {T}_{4+1} = {}^{7}{C}_{4}{(\frac{x}{3})}^{7-4}{(-3y)}^{4} \\ \\ \tt \implies {T}_{5} = \frac{7!}{(7-4)!4!}{(\frac{x}{3})}^{3}{(-3y)}^{4} \\ \\ \tt \implies {T}_{5} = \frac{7×6×5}{1×2×3}(\frac{{x}^{3}}{27})(81{y}^{4}) \\ \\ \tt {T}_{5} = 35×3{x}^{3}{y}^{4} \\ \\ \tt \implies {T}_{5} = 105{x}^{3}{y}^{4} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt {T}_{5} = 105{x}^{3}{y}^{4} }}}} \\ \\ \tt \therefore \boxed{\bold{\underline{\purple{\tt Correct\:option\:is\:option\:1)\:105}}}}$