If the second, third and fourth terms in the | vedclass

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Question

$ { }^n C_1 x^{n-1} y=135 $ ...........($i$)

$ { }^n C_2 x^{n-2} y^2=30 $ ............($ii$)

$ { }^n C_3 x^{n-3} y^3=\frac{10}{3} $ ............

Vedclass · Accepted Answer

$806$

Vedclass · Answer

$ { }^n C_1 x^{n-1} y=135 $ ...........($i$)

$ { }^n C_2 x^{n-2} y^2=30 $ ............($ii$)

$ { }^n C_3 x^{n-3} y^3=\frac{10}{3} $ ............($iii$)

$ 	ext { By } \frac{(i)}{(i i)} $

$ \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 $ ...............($v$)

$ 	ext { By } \frac{(	ext { iv) }}{(v)} $

$ \frac{{ }^n C_1 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$

put in ($v$)

$ \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$

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

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