Question

The 2nd, 3rd and 4th terms in the expansion of ${(x+y)}^n$ are $240,720,1080$ respectively; find $x,y,n$.

Answer 1

General term $T_{r+1}$ of $(x+y{)}^n$ is given by

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$⟹T_2={}^n{C}_1\cdot x^{n-1}\cdot y=240$ ..... $\left(A\right)$

$T_3={}^n{C}_2\cdot x^{n-2}\cdot y^2=720$

$T_4={}^n{C}_3\cdot x^{n-3}\cdot y^3=1080$

$\frac{T_3}{T_2}=\frac{(n-1)}{2}\times \frac{y}{x}=3$ ....... $\left(i\right)$

$\frac{T_4}{T_2}=\frac{(n-1)(n-2)}{3\times 2}\times \frac{x^2}{y^2}=\frac{9}{2}$ ........ $\left(ii\right)$

$\frac{T_4}{T_3}=\frac{(n-2)}{3}\times \frac{y}{x}=\frac{3}{2}$ ....... $\left(iii\right)$

Dividing equation $\left(i\right)$ by equation $\left(iii\right):-$

$\frac{n-1}{2}\times \frac{3}{n-2}=2$

$\Rightarrow 3n-3=4n-8$

$\Rightarrow 5=n$

$\Rightarrow \frac{y}{x}=\frac{3}{2}$ ..... From $\left(iii\right)$

Putting $y=\frac{3}{2}x$

${5x}^5\times \frac{3}{2}=240$ ........ From $\left(A\right)$

$x^5=32$

$x=2,y=3,n=5$