If the coefficients of \[\left( 2r + 4 \right)\text{ th and } \left( r - 2 \right)\] th terms in the expansion of \[\left( 1 + x \right)^{18}\] are equal, find *r*.

#### Solution

\[Given: \]

\[(1 + x )^{18} \]

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

\[\text{ Therefore, the coefficients of the (2r + 4)th and (r - 2)th terms in the given expansion are } ^{18}{}{C}_{2r + 4 - 1} \text{ and } ^{18}{}{C}_{r - 2 - 1} \]

\[\text{ For these coefficients to be equal, we must have } \]

\[^{18}{}{C}_{2r + 3} =^{18}{}{C}_{r - 3} \]

\[ \Rightarrow 2r + 3 = r - 3 or, 2r + 3 + r - 3 = 18 [ \because ^{n}{}{C}_r = ^{n}{}{C}_s \Rightarrow r = s \text{ or } r + s = n]\]

\[ \Rightarrow r = - 6 \text{ or } , r = 6\]

\[\text{ Neglecting negative value We get} \]

\[r = 6\]