If the second, third and fourth terms in the expansion of $(x+y)^n$ are 135,30 and $\frac{10}{3}$, respectively, then $6\left(n^3+x^2+y\right)$ is equal to _______

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Aicharya · Accepted Answer

${ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \mathrm{y}=135$....(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) $\begin{aligned} & 	ext { By } \frac{(i)}{(	ext { ii) }} \ & \frac{{ }^n C_1}{{ }^n C_2} \frac{x}{y}=\frac{9}{2}...(iv) \ & 	ext { By } \frac{(	ext { ii) }}{(	ext { iii) }} \ & \frac{{ }^n C_2}{{ }^n C_3} \frac{x}{y}=9....(v)\end{aligned}$ $\begin{aligned} & 	ext { By } \frac{(	ext { iv })}{(v)} \ & \frac{{ }^n C_1{ }^n C_3}{{ }^n C_2{ }^n C_2}=\frac{1}{2} \ & \frac{2 n^2(n-1)(n-2)}{6}=\frac{n(n-1)}{2} \frac{n(n-1)}{2} \ & 4 n-8=3 n-3 \ & \Rightarrow n=5 \ & 	ext { put in (v) }\end{aligned}$ $\begin{aligned} & \frac{x}{y}=9 \ & x=9 y \ & 	ext { put in (i) } \ & { }^5 C_1 x^4\left(\frac{x}{9}ight)=135 \ & x^5=27 	imes 9 \ & \Rightarrow x=3, \quad y=\frac{1}{3} \ & 6\left(n^3+x^2+yight) \ & =6\left(125+9+\frac{1}{3}ight) \ & =806\end{aligned}$

If the second, third and fourth terms in the expansion of $(x+y)^n$ are 135,30 and $\frac{10}{3}$,…

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