Question

If for some positive integer $n$, the coefficients of three consecutive terms in the binomial expansion of $(1+x{)}^{n+5}$ are in the ratio $5:10:14,$ then the largest coefficient in this expansion is

• A. $792$
• B. $252$
• C. $462$
• D. $330$

Answer 1

The correct option is C $462$

Given: $(1+x{)}^{n+5}$
The general term, $T_{r+1}{=}^{n+5}{C}_r{x}^r$
Let three consecutive terms be $T_r,T_{r+1}$ and $T_{r+2}.$
The, coefficients of three consecutive terms ${}^{n+5}{C}_{r-1},{}^{n+5}{C}_r,{}^{n+5}{C}_{r+1}$
Now,
${}^{n+5}{C}_{r-1}:{}^{n+5}{C}_r:{}^{n+5}{C}_{r+1}=5:10:14$
Take
$\frac{{}^{n+5}{C}_{r-1}}{{}^{n+5}{C}_r}=\frac{5}{10}$
$\Rightarrow \frac{\frac{(n+5)!}{(r-1)!(n+6-r)!}}{\frac{(n+5)!}{r!(n+5-r)!)}}=\frac{1}{2}$
$\Rightarrow \frac{r}{n+6-r}=\frac{1}{2}$
$\Rightarrow n+6-r=2r$
$\Rightarrow n+6=3r\cdots \left(1\right)$
Take
$\frac{{}^{n+5}{C}_r}{{}^{n+5}{C}_{r+1}}=\frac{10}{14}$
$\Rightarrow \frac{\frac{(n+5)!}{\left(r\right)!(n+5-r)!}}{\frac{(n+5)!}{(r+1)!(n+4-r)!)}}=\frac{5}{7}$
$\Rightarrow \frac{r+1}{n+5-r}=\frac{5}{7}$
$\Rightarrow 5n+25-5r=7r+7$
$\Rightarrow 5n+18=12r\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$5n+18=4n+24$
$\Rightarrow n=6$ and $r=4$

Now, largest coefficient in the expansion $(1+x{)}^{11}$
$={}^{11}{C}_6={}^{11}{C}_5=462$