Question

For a positive integer $n$, ${\left(1+\frac{1}{x}\right)}^n$ is expanded in increasing powers of $x$.

If three consecutive coefficients in this expansion are in the ratio, $2:5:12$, then $n$ is equal to

Answer 1

Step 1.Find the value of $\mathit{r}$:

Let three consecutive coefficients are ${}^{n}C_{r-1}:{}^{n}C_r:{}^{n}C_{r+1}::2:5:12$

$\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{2}{5}$

$\Rightarrow$ $\frac{{\displaystyle \raisebox{1ex}{$n!$}\!\left/ \!\raisebox{-1ex}{$\left(n-r+1\right)$}\right.}!\left(r-1\right)!}{{\displaystyle \raisebox{1ex}{$n!$}\!\left/ \!\raisebox{-1ex}{$\left(n-r\right)!r!$}\right.}}=\frac{2}{5}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$\Rightarrow$ $\frac{\left(n-r\right)!r!}{\left(n-r+1\right)!\left(r-1\right)!}=\frac{2}{5}$

$\Rightarrow$$\frac{\left(n-r\right)!r\left(r-1\right)!}{\left(n-r+1\right)\left(n-r\right)!\left(r-1\right)!}=\frac{2}{5}$

$\Rightarrow$ $\frac{r}{(n-r+1)}=\frac{2}{5}$

$\Rightarrow$ $7r=2n+2$

$\Rightarrow$ $r=\frac{(2n+2)}{7}$ …..(1)

and $\frac{{}^{n}C_r}{{}^{n}C_{r+1}}=\frac{5}{12}$

$\Rightarrow$$\frac{{\displaystyle \raisebox{1ex}{$n!$}\!\left/ \!\raisebox{-1ex}{$\left(n-r\right)!r!$}\right.}}{{\displaystyle \raisebox{1ex}{$n!$}\!\left/ \!\raisebox{-1ex}{$\left(n-r-1\right)!\left(r+1\right)!$}\right.}}=\frac{5}{12}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$\Rightarrow$ $\frac{\left(n-r-1\right)!\left(r+1\right)!}{\left(n-r\right)!r!}=\frac{5}{12}$

$\Rightarrow$ $\frac{\left(n-r-1\right)!\left(r+1\right)r!}{\left(n-r\right)\left(n-r-1\right)!r!}=\frac{5}{12}$

$\Rightarrow$ $\frac{(r+1)}{(n-r)}=\frac{5}{12}$

$\Rightarrow$ $17r=5n-12$

$\Rightarrow$ $r=\frac{(5n-12)}{17}$ ……(2)

Step 2. Equating equation (1) and (2):

$\Rightarrow$$\frac{(2n+2)}{7}=\frac{(5n-12)}{17}$

$\Rightarrow$$34n+34=35n-84$

Hence, $\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{118}$