Question

In the binomial expansion of $(1+x{)}^{15}$ the coefficients of $x^r$ and $x^{r+3}$ are equal Then $r$ is _______

• A. 4
• B. 6
• C. 8
• D. 7

Answer 1

Answer: (B)

Given, $(1+x{)}^{15}$
Now, $T_{r+1}={}^{15}{C}_r{x}^r$
and $T_{(r+3)+1}={}^{15}{C}_{r+3}{x}^{r+3}$
According to question
coefficient of $x^r=$ coefficient of $x^{r+3}$
$\Rightarrow {}^{15}{C}_r={}^{15}{C}_{r+3}$
$\Rightarrow \frac{15!}{r!(15-r)!}=\frac{15!}{(r+3)!(12-r)!}$
$\Rightarrow \frac{1}{(15-r)(14-r)(13-r)}$
$=\frac{1}{(r+3)(r+2)(r+1)}$
$\Rightarrow (r+1)(r+2)(r+3)=(15-r)$
$\Rightarrow \left(r^2+3r+2\right)(r+3)=\left(210-29r+r^2\right)$
$\Rightarrow r^3+3r^2+2r+3r^2+9r+6$
$=2930-377r+13r^2-210r+29r^2-r^3$
$\Rightarrow 2r^3-36r^2+598r-2924=0$
$\Rightarrow r^3-18r^2+299r-1462=0$
$\Rightarrow (r-6)\left(r^2-12r+227\right)=0$
$\Rightarrow r=6$ and $r^2-12r+227=0$ gives imaginary roots.
Alternate Method
${}^{15}{C}_r={}^{15}{C}_{r+3}$
$\Rightarrow r+(r+3)=15$
$\left(\because {}^n{C}_x={}^n{C}_y\right)$
$\Rightarrow 2r+3=15$
$\Rightarrow x+y=n$
$\Rightarrow 2r=12$
$\Rightarrow r=6$