Question

If the coefficient of $(2r+4{)}^{th}$ term and $(r-2{)}^{th}$ term in the expansion of $(1+x{)}^{18}$ are equal then $r$

• A. $9$
• B. $4$
• C. $6$
• D. $3$

Answer 1

The correct option is C $6$

In the expansion of $(1+x{)}^{18}$

$T_{r+1}{=}^{18}{C}_r{1}^{18-r}{x}^r$

$\therefore T_{2r+4}{=}^{18}{C}_{2r+3}{1}^{18-2r-3}{x}^{2r+3}$

and $T_{r-2}{=}^{18}{C}_{r-3}{1}^{18-r+3}{x}^{r-3}$

$T_{2r+4}=T_{r-2}$

i.e.${}^{18}{C}_{2r+3}{=}^{18}{C}_{r-3}$

$\frac{18!}{(18-2r-3)!(2r+3)!}=\frac{18!}{(18-r+3)!(r-3)!}$

$(18-2r-3)!(2r+3)!=(18-r+3)!(r-3)!$

$(15-2r)!(2r+3)!=(21-r)!(r-3)!$

Now, Posibilities are $15-2r=21-r\Rightarrow r=-6$

But we know $r$ is a an integer such that $rϵ[0,18]$

Now, $15-2r=r-3\Rightarrow 3r=18\Rightarrow r=6$

Checking with $2r+3=21-r\Rightarrow 3r=18\Rightarrow r=6$