Question

The coefficients of three consecutive terms of $(1 + x)^n+5$ are in the ratio $5:10:14$. Then, $n$ is equal to

Answer 1

Let the three consecutive terms in $(1+x)^{n+5}$ be $t_{r}, t_{r+1}, t_{r+2} 2$ having coefficients
${ }^{n+5} C_{r-1},{ }^{n+5} C_{r},{ }^{n+5} C_{r+1}$.
Given, ${ }^{n+5} C_{r-1}:{ }^{n+5} C_{r}:{ }^{n+5} C_{r+1}=5: 10: 14$
$\therefore \frac{{ }^{n+5} C_{r}}{{ }^{n+5} C_{r-1}}=\frac{10}{5}$ and $\frac{{ }^{n+5} C_{r+1}}{{ }^{n+5} C_{r}}=\frac{14}{10}$
$\Rightarrow \frac{n+5-(r-1)}{r}=2$ and $\frac{n+r+5}{r+1}=\frac{7}{5}$
$\Rightarrow n-r+6=2$ and $5 n-5 r+25=7 r+7$
$\Rightarrow n+6=3 r$ and $5 n+18=12 r$
$\therefore \frac{n+6}{3}=\frac{5 n+18}{12}$
$\Rightarrow 4 n+24=5 n+18 \Rightarrow n=6$