Question

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

• A. n = 4, x = 2, y = 3
• B. n = 5, x = 2, y = 3
• C. n = 4, x = 3, y = 3
• D. n = 6, x = 2, y = 3

Answer 1

The correct option is B

n = 5, x = 2, y = 3

Given $T_2{=}^n{C}_1{x}^{n-1}y=240$ . . .(i)
$T_3{=}^n{C}_2{x}^{n-2}{y}^2=720$ . . .(ii)
and $T_4{=}^n{C}_3{x}^{n-3}{y}^3=1080$
Multiplying (i) and (iii), we get
${}^n{C}_1{\times }^n{C}_3\times x^{2n-4}.y^4=240\times 1080\Rightarrow {}^n{C}_1{\times }^n{C}_3\times (x^{n-2}{y}^2{)}^2=240\times 1080\Rightarrow {}^n{C}_1{\times }^n{C}_3\times {\left(\frac{720}{{}^n{C}_2}\right)}^2=240\times 1080$
$\Rightarrow \frac{n\times n(n-1)(n-2)}{\mathrm{1.2.3}}\times \frac{720\times 720}{\frac{n^2(n-1{)}^2}{4}}=240\times 1080\Rightarrow \frac{n^2(n-2)(n-1)\times 2\times 720\times 720}{3n^2(n-1{)}^2}=240\times 1080\Rightarrow \frac{4(n-2)}{3(n-1)}=1\therefore n=5$
from (i), ${}^5{C}_1{x}^{4}y=240$
$\Rightarrow x^{4}y=48$
from (ii), ${}^5{C}_2{x}^3{y}^2=720$
$x^3{y}^2=72$
$\Rightarrow x^3{\left(\frac{48}{x^4}\right)}^2=72\Rightarrow (48{)}^2=72x^5\therefore x^5=32\therefore x=2$
from (iv), $2^{4}y=48$
$\therefore$ y = 3
Hence $n=5,x=2,y=3$