Question

The coefficients of three consecutive terms of $(1+x{)}^{n+5}$ are in the ratio $5:10:14$, then $n=$

Answer 1

$T_{r+1}{=}^{n+5}{C}_r{x}^r$

$3$ consecutive terms are $T_r,T_{r+1},T_{r+2}$

$\therefore$Co efficients, $T_r{=}^{n+5}{C}_{r-1}$

$T_{r+1}{=}^{n+5}{C}_r$

$T_{r+2}{=}^{n+5}{C}_{r+1}$

A.T.P

$\frac{T_{r+1}}{T_r}=\frac{10}{5}=\frac{(n+5)!}{r!(n+5-r)!}\times \frac{(r-1)!(n+6-r)!}{(n+5)!}$

$\Rightarrow 2=\frac{n+6-r}{r}$

$\therefore 3r=n+6\to 1$

$\frac{T_{r+2}}{T_r}=\frac{14}{10}=\frac{(n+5)!}{(r+1)!(n+4-r)!}\times \frac{r!(n+5-r)!}{(n+5)!}$

$\Rightarrow \frac{7}{5}=\frac{n+5-r}{r+1}$

$\Rightarrow 7r+7=5n+25-5r$

$\Rightarrow 12r=5n+18\to 2$

$\therefore 4n+24=5n+18$ (From $112r=4n+24$)

$\Rightarrow n=6$