Question

The second,third and fourth terms in the expansion of $(x+a{)}^n$ are $240$ , $720$ , $1080$ respectively.Find the value of $x$ , $a$ and $n$.

Answer 1

${}^n{C}_1{x}^{n-1}a=240$ ……$\left(1\right)$

${}^n{C}_2{x}^{n-2}{a}^2=720$ …….$\left(2\right)$

${}^n{C}_3{x}^{n-3}{a}^3=1080$ ……….$\left(3\right)$

$\frac{\left(1\right)}{\left(2\right)}$ $\Rightarrow \frac{{}^n{C}_1}{{}^n{C}_2}\frac{x}{a}=\frac{1}{3}$ ………$\left(4\right)$

$\frac{\left(2\right)}{\left(3\right)}$ $\Rightarrow \frac{x}{a}\frac{{}^n{C}_2}{{}^n{C}_3}=\frac{2}{3}$ …….$\left(5\right)$

$\frac{\left(4\right)}{\left(5\right)}$ $\Rightarrow \frac{{}^n{C}_1\times {}^n{C}_3}{{}^n{C}_2\times {}^n{C}_2}=\frac{1}{3}\times \frac{3}{2}$

$\Rightarrow \frac{(n-2)!(n-2)!2!2!}{(n-1)!(n-3)!3!1!}=\frac{1}{2}$

$\Rightarrow \frac{4(n-2)}{(n-1)6}=\frac{1}{2}$

$3(n-1)=4(n-2)$

$5=n$

$\Rightarrow \frac{{}^5{C}_1}{{}^5{C}_2}\frac{x}{a}=\frac{1}{3}$

$\Rightarrow \frac{2!3!}{4!}\frac{n}{a}=\frac{1}{3}$

$3n=2a$

$\left(1\right)\Rightarrow {}^5{C}_1{x}^4\left(\frac{3}{2}x\right)=240$

$x^5=16\times 2$

$x=2$

$\left(4\right)$ $\Rightarrow \frac{{}^5{C}_1}{{}^5{C}_2}\frac{2}{a}=\frac{1}{3}$

$\frac{1}{2}\times \frac{2}{a}=\frac{1}{3}$

$a=3$.