If the second, third and fourth terms in the expansion of (x+y)^n are 135, 30 and [(10)/(3…

Question

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 \_\_\_\_\_\_\_\_\_\_.

Answer: 806

Accepted Answer

$\begin{gathered} {T}_{2}={}^{n}{C}_{1}{y}^{1}\cdot {x}^{n-1}=135 \ {T}_{3}={}^{n}{C}_{2}{y}^{2}\cdot {x}^{n-2}=30 \ {T}_{4}={}^{n}{C}_{3}{y}^{3}{x}^{n-3}=\frac{10}{3} \end{gathered}$

$\begin{gathered} \Rightarrow \frac{135}{30} =(\frac{x}{y})\frac{n\cdot 2}{n(n-1)}=(\frac{2}{n-1})(\frac{x}{y})\, \ \frac{30}{\frac{10}{3}} =\frac{n(n-1)}{2}\frac{3!}{n(n-1)(n-2)}(\frac{x}{y}) \ 9 =(\frac{3}{n-2})(\frac{x}{y}) $

$ \Rightarrow 3(n-2)=\frac{135}{60}(n-1)\Rightarrow n=5 \ \Rightarrow x=9y\, \ y\cdot {x}^{4}=27\Rightarrow \frac{x}{9}\cdot {x}^{4}={3}^{3} \ \Rightarrow {x}^{5}={3}^{5}\Rightarrow x=3y=\frac{1}{3} \ \Rightarrow 6({5}^{3}+{3}^{2}+\frac{1}{3})=6(125+9+\frac{1}{3}) \end{gathered}$

$=6(134)+2=806$

Solutions · 1