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

Answer: 6

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 $5n-5r+25=7r+7$
$\Rightarrow n+6=3r$ and $5n+18=12r$
$\therefore \frac{n+6}{3}=\frac{5n+18}{12}$
$\Rightarrow 4n+24=5n+18\Rightarrow n=6$