Question

In the coefficients of (2r+4)th and (r-2)th terms in the expansion of $(1=x{)}^{18}$ are equal, find r.

Answer 1

We know that the coefficient of rth term in the expansion of $(1+x{)}^{n}is{}^n{C}_{r-1}$

$\therefore$ Coefficient of (2r+4)th term of the expansion $(1+x{)}^{18}={}^{18}{C}_{2r+4-1}={}^{18}{C}_{2r+3}$ and, coefficient of (r-2)th term of the expansion $(1+x{)}^{18}={}^{18}{C}_r-2-1={}^{18}{C}_{r-3}$

It is given that these coefficients are equal.

$\therefore {}^{18}{C}_{2r+3}={}^{18}{C}_{r-3}$

$\Rightarrow 2r+3=r-3$ or, 2r+3 +r -3 =18

$\Rightarrow$ r =-6 or, 3r =18

$\left[\begin{array}{c}\because {}^n{C}_r={}_s^C\\ \Rightarrow r=sor,r+s=n\end{array}\right]$

$\Rightarrow r=-6$ or, r=6

$\Rightarrow r=6[\because$ r=-6 is not possible]