Question

Find a,b and n in the expansion of $(x+b{)}^n$ if the first three terms in the expansion are 729, 7290 and 30375 respectively.

Answer 1

It is given that

$T_1=729$

$T_2=7290$

and, $T_3=30375$

$\therefore T_1={}^n{C}_0\times a^n=729$

$T_2={}^n{C}_1\times a^{n-1}\times b=7290$

$\frac{T_2}{T_1}=\frac{{}^n{C}_{n-1}\times a^{n-1}\times b}{{}^n{C}_0\times a^n}=\frac{7290}{729}$

$\Rightarrow \frac{{}^n{C}_{n-1}\times a^{n-1}\times b}{{}^n{C}_0\times a^n}=10$

$\Rightarrow \frac{{}^n{C}_{n-1}}{1}\times \frac{b}{a}=10$

$\Rightarrow \frac{n!}{(n-n+1)!(n-1)!}\times \frac{b}{a}=10$

$\Rightarrow \frac{n!}{(n-1)!}\times \frac{b}{a}=10$

$\Rightarrow \frac{n(n-1)!}{(n-1)!}\times \frac{b}{a}=10$

$\Rightarrow \frac{b}{a}=\frac{10}{n}......\left(i\right)$

and,

$\frac{T_3}{T_2}=\frac{{}^n{C}_{n-2}\times a^{n-2}\times b^2}{{}^n{C}_{n-1}\times a^{n-1}\times b}=\frac{30375}{7290}$

$\Rightarrow \frac{{}^n{C}_{n-2}}{{}^n{C}_{n-1}}\times \frac{b}{a}=\frac{25}{6}$

$\Rightarrow \frac{n-2+1}{n-(n-1)+1}\times \frac{b}{a}=\frac{26}{6}$

$[\because \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}]$

$\Rightarrow \frac{n-1}{2}\times \frac{b}{x}=\frac{25}{6}$

$\Rightarrow \frac{b}{a}=\frac{25}{6}\times \frac{2}{(n-1)}$

$\Rightarrow \frac{b}{a}=\frac{25}{3(n-1)}$

Comparing equation (i)and equation (ii), we get

$\frac{10}{n}=\frac{25}{3(n-1)}$

$\Rightarrow 30(n-1)=25$

$\Rightarrow 30n-30=25n$

$\Rightarrow 5n=30$

$\Rightarrow n=6$

Now,

$T_1={}^n{C}_0\times a^n=729$

$\Rightarrow a^n=729$

$\Rightarrow a^6=729$

$\Rightarrow a^6=3^6[\because n=6]$

$\Rightarrow a=3$

Putting a =3 in n =6 in equation (i), we get

$\frac{b}{3}=\frac{10}{6}$

$\Rightarrow b=\frac{10}{2}=5$

Hence, a =3, b =5 and n =6,