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 $x^{r+3}$
$\Rightarrow { }^{15} C_{r}={ }^{15} C_{r+3}$
$\Rightarrow \frac{15 !}{r !(15-r) !}=\frac{15 !}{(r+3) !(12-r) !}$
$\Rightarrow \frac{1}{(15-r)(14-r)(13-r)}$
$=\frac{1}{(r+3)(r+2)(r+1)}$
$\Rightarrow (r+1)(r+2)(r+3)=(15-r)$
$\Rightarrow \left(r^{2}+3 r+2\right)(r+3)=\left(210-29 r+ r^{2}\right)$
$\Rightarrow r^{3}+3 r^{2}+2 r+3 r^{2}+9 r+6$
$=2930-377 r+13 r^{2}-210 r+29 r^{2}-r^{3}$
$\Rightarrow 2 r^{3}-36 r^{2}+598 r-2924=0$
$\Rightarrow r^{3}-18 r^{2}+299 r-1462=0$
$\Rightarrow (r-6)\left(r^{2}-12 r+227\right)=0$
$\Rightarrow r=6$ and $r^{2}-12 r+227=0$ gives imaginary roots.
Alternate Method
${ }^{15} C_{r}={ }^{15} C_{r+3}$
$\Rightarrow r+(r+3)=15$
$\left(\because{ }^{n} C_{x}={ }^{n} C_{y}\right)$
$\Rightarrow 2 r+3=15$
$\Rightarrow x +y =n$
$\Rightarrow 2 r =12$
$\Rightarrow r=6$