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(B) The terms in the expansion of $(x+y)^n$ are given by $T_{r+1} = {}^nC_r x^{n-r} y^r$.Given:$T_2 = {}^nC_1 x^{n-1} y = 135$ ...........$(i)$$T_3 =

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(B) The terms in the expansion of $(x+y)^n$ are given by $T_{r+1} = {}^nC_r x^{n-r} y^r$.Given:$T_2 = {}^nC_1 x^{n-1} y = 135$ ...........$(i)$$T_3 = {}^nC_2 x^{n-2} y^2 = 30$ ............$(ii)$$T_4 = {}^nC_3 x^{n-3} y^3 = \frac{10}{3}$ ............$(iii)$Dividing $(i)$ by $(ii)$:$\frac{{}^nC_1 x^{n-1} y}{{}^nC_2 x^{n-2} y^2} = \frac{135}{30}$ $\Rightarrow \frac{n}{\frac{n(n-1)}{2}} \cdot \frac{x}{y} = \frac{9}{2}$ $\Rightarrow \frac{2}{n-1} \cdot \frac{x}{y} = \frac{9}{2}$ $\Rightarrow \frac{x}{y} = \frac{9(n-1)}{4}$ ............$(iv)$Dividing $(ii)$ by $(iii)$:$\frac{{}^nC_2 x^{n-2} y^2}{{}^nC_3 x^{n-3} y^3} = \frac{30}{10/3}$ $\Rightarrow \frac{\frac{n(n-1)}{2}}{\frac{n(n-1)(n-2)}{6}} \cdot \frac{x}{y} = 9$ $\Rightarrow \frac{3}{n-2} \cdot \frac{x}{y} = 9$ $\Rightarrow \frac{x}{y} = 3(n-2)$ ............$(v)$Equating $(iv)$ and $(v)$:$\frac{9(n-1)}{4} = 3(n-2)$ $\Rightarrow 3(n-1) = 4(n-2)$ $\Rightarrow 3n-3 = 4n-8$ $\Rightarrow n = 5$.Substitute $n=5$ into $(v)$:$\frac{x}{y} = 3(5-2) = 9 \Rightarrow x = 9y$.Substitute $n=5$ and $x=9y$ into $(i)$:${}^5C_1 (9y)^4 y = 135$ $\Rightarrow 5 \cdot 6561 y^5 = 135$ $\Rightarrow y^5 = \frac{135}{5 \cdot 6561} = \frac{27}{6561} = \frac{1}{243} = (\frac{1}{3})^5$ $\Rightarrow y = \frac{1}{3}$.Then $x = 9(\frac{1}{3}) = 3$.Calculate $6(n^3+x^2+y)$:$6(5^3 + 3^2 + \frac{1}{3}) = 6(125 + 9 + \frac{1}{3}) = 6(134 + \frac{1}{3}) = 804 + 2 = 806$.

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

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