The coefficients of three successive terms in the expansion of ${(1+x)}^n$  are 165, 330 and 462 respectively, then the value of $n$  will be

Given expansion  ${(1+x)}^n$ is

We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$ be the expansion of ${(x+a)}^n)$

Let the three successive terms be $T_r,T_{r+1}$  and $T_{r+2}$ , then

$r^{th}$  term:

$T_{\left(r-1\right)+1}={}^n{\text{C}}_{r-1}{x}^{n-r+1}{\left(a\right)}^{r-1}$

Coefficient of $r^{th}$  term is  ${}^n{C}_{r-1}$

$\because {}^n{C}_{r-1}=\mathrm{165..................}\left(1\right)$      

${(r+1)}^{th}$  term:

$T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$

Coefficient of ${(r+1)}^{th}$  term is  ${}^n{C}_r$

$\because {}^n{C}_r=\mathrm{330..................}\left(2\right)$

${(r+2)}^{th}$  term:

$T_{\left(r+1\right)+1}={}^n{\text{C}}_{r+1}{x}^{n-r-1}{\left(a\right)}^{r+1}$  

Coefficient of ${(r+2)}^{th}$  term is  ${}^n{C}_{r+1}$

${\because }^n{C}_{r+1}=\mathrm{462.........................}\left(3\right)$

Dividing Eqn (2)by Eqn (1), we get

$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{330}{165}$ $\left(\therefore \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}\right)$

          $\frac{n-r+1}{r}=2$            …(4)

 Dividing Eqn (3)by Eqn (2), we get

 $\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{462}{330}$ $\left(\therefore \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}\right)$

         $\frac{n-r}{r+1}=\frac{7}{5}$                  ….(5)

From (4), $n=3r-1$  and from (5)

                           $n=\frac{12r+7}{5}$

$\Rightarrow 3r-1=\frac{12r+7}{5}$  

$\begin{array}{c}\Rightarrow 3r=12\\ \end{array}$

$\because r=4$  in  Eqn (5) we get $n=11$