Question

If the coefficients of $(2r+4{)}^{th}$ and $(r-2{)}^{th}$ terms in the expansion of $(1+x{)}^{18}$ are equal, then the value of $r$ is

Answer 1

In the expansion of $(1+x{)}^{18}$, we have $T_{p+1}={}^{18}{C}_p{x}^p$
$\therefore T_{(2r+4)}=T_{(2r+3)+1}={}^{18}{C}_{2r+3}.x^{2r+3}\cdots \left(1\right)$
and $T_{(r-2)}=T_{(r-3)+1}={}^{18}{C}_{(r-3)}.x^{r-3}\cdots \left(2\right)$
It is given that the coefficients of $T_{(2r+4)}$ and $T_{(r-2)}$ are equal.
So, from $\left(1\right)$ and $\left(2\right)$, we get
${}^{18}{C}_{2r+3}={}^{18}{C}_{r-3}$
$\Rightarrow (2r+3)=(r-3)$ or $(2r+3)+(r-3)=18$ $[\because {}^n{C}_p{=}^n{C}_q\Rightarrow p=q\text{or}p+q=n]$
Now, $(2r+3)=(r-3)$ or $(2r+3)+(r-3)=18$
$\Rightarrow r=-6$ or $r=6$
$\Rightarrow r=6$
[$\because$ the negative value of $r$ is not permissible].
Hence, $r=6$