Question

The coefficient of $3$ consecutive terms in the expansion of ${(1+x)}^n$ are in the ratio $3:8:14$. Find $n$.

Answer 1

In general term in expansion ${\left(1+x\right)}^n$ is given by $t_r+1=n{c}_r{x}^r$

so there is consecutive terms will have coefficient as

$n{c}_{r-1},n{c}_{r}andn{c}_{r+1}$

given,

$\frac{n{c}_{r-1}}{n{c}_r}=\frac{3}{8}and\frac{n{c}_r}{n{c}_{r+1}}=\frac{8}{14}$

$\begin{array}{c}\Rightarrow \frac{\frac{n!}{\left(r-1\right)\left(n-r+1\right)!}}{\frac{n!}{\left(r\right)!\left(n-r\right)!}}=\frac{3}{8}\\ \\ \Rightarrow \frac{\left(r\right)!\left(n-r\right)!}{\left(r-1\right)!\left(n-r+1\right)!}=\frac{r}{n-r+1}=\frac{3}{8}\\ \Rightarrow 8r=3n-3r+3\\ \Rightarrow 11r=3n+\mathrm{3............}\left(1\right)\end{array}$

Also $\frac{n{c}_r}{n{c}_{r+1}}=\frac{8}{14}$

$\begin{array}{c}\Rightarrow \frac{\frac{n!}{\left(r\right)!\left(n-r\right)!}}{\frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}}=\frac{8}{14}\\ \Rightarrow \frac{\left(r+1\right)!\left(n-r-1\right)!}{\left(r\right)!\left(n-r\right)!}=\frac{\left(r+1\right)}{n-r}=\frac{8}{14}\\ \Rightarrow 14r+14=8n-8r\\ \Rightarrow 22r=8n-14\\ \Rightarrow 11r=4n-\mathrm{7............}\left(2\right)\end{array}$

From $1$ and $2$ we get

$\begin{array}{c}4n-7=3n+3\\ \Rightarrow n=10\end{array}$