Mathematics - Binomial Theorem Question with Solution | TestHub

${}^{\mathrm{n}}{\mathrm{C}}_1{\mathrm{x}}^{\mathrm{n}-1}\mathrm{y}=135$....(i) ${}^n{C}_2{x}^{n-2}{y}^2=30$....(ii) ${}^{\mathrm{n}}{\mathrm{C}}_3{\mathrm{x}}^{\mathrm{n}-3}{\mathrm{y}}^3=\frac{10}{3}$....(iii) $\begin{array}{rl}& \text{ By }\frac{(i)}{(\text{ ii) }}\\ & \frac{{}^n{C}_1}{{}^n{C}_2}\frac{x}{y}=\frac{9}{2}...(iv)\\ & \text{ By }\frac{(\text{ ii) }}{(\text{ iii) }}\\ & \frac{{}^n{C}_2}{{}^n{C}_3}\frac{x}{y}=\mathrm{9....}(v)\end{array}$ $\begin{array}{rl}& \text{ By }\frac{(\text{ iv })}{(v)}\\ & \frac{{}^n{C}_1{}^n{C}_3}{{}^n{C}_2{}^n{C}_2}=\frac{1}{2}\\ & \frac{2n^2(n-1)(n-2)}{6}=\frac{n(n-1)}{2}\frac{n(n-1)}{2}\\ & 4n-8=3n-3\\ & \Rightarrow n=5\\ & \text{ put in (v) }\end{array}$ $\begin{array}{rl}& \frac{x}{y}=9\\ & x=9y\\ & \text{ put in (i) }\\ & {}^5{C}_1{x}^4\left(\frac{x}{9}\right)=135\\ & x^5=27\times 9\\ & \Rightarrow x=3,y=\frac{1}{3}\\ & 6\left(n^3+x^2+y\right)\\ & =6\left(125+9+\frac{1}{3}\right)\\ & =806\end{array}$