Question

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

Answer 1

Second, third, fourth, terms in $(x+a{)}^n$ are

$240,720$ & $1080$. Find $x,a,n$

$\to$ binomial expansion $(x+a{)}^n$

$T_{n+1}={}^n{C}_2{x}^{x+2}{y}^2$

$\to$ for second term $r=1$

$T_2={}^n{C}_1(x{)}^{n+1}.y$

$\therefore 240={}^n{C}_1(x{)}^{n-1}.a....(i)$

$\to$ for third term $r=2$

$T_3={}^n{C}_2{x}^{n-2}.a^2$

$\therefore 720={}^n{C}_2{\alpha }^{n-2}{a}^2....\left(ii\right)$

$\to$ for fourth term $r=3$

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

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

$\to$ equation $\left(ii\right)$ divide by equation $\left(i\right)$

$\therefore \frac{720}{240}=\frac{{}^n{C}_2{x}^{n-2}{a}^2}{{}^n{C}_1{x}^{n-1}a}$

$\therefore 3=\frac{n-1}{2}\times \frac{a}{x}....\left(iv\right)$

$\to$ equation $\left(iii\right)$ divide by equation $\left(ii\right)$

$\therefore \frac{1080}{720}=\frac{{}^n{C}_3{x}^{n-3}{a}^3}{{}^n{C}_2{x}^{n-2}{a}^2}$

$\therefore \frac{3}{2}=\frac{(n-2)}{3}\frac{a}{x}....\left(v\right)$

equation $\left(iv\right)$ divide by equation $\left(v\right)$

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

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

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

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

$\therefore n=-3+8$

$\therefore n=6$

Putting $n$ is equation $\left(iv\right)$ & $\left(v\right)$

$3=e\times \frac{a}{x}\to \left(vi\right)$

$\frac{3}{2}=\frac{a}{x}.....\left(vii\right)$

$\to$ putting $n$ & $\frac{a}{x}$ value in equation $\left(i\right)$

$\therefore 240={}^5{C}_1\frac{x^n}{x}.a$

$\therefore 240={}^5{C}_2{x}^n.\frac{3}{2}$

$\therefore x^n=\frac{5\times 3}{2\times 240}\frac{240\times 2}{5\times 3}$

$\therefore x^5=32\therefore x^5=2^5$

$\therefore x=2$ & $x=3$