Question

If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)²ⁿ are in A.P., show that  \[2 n^2 - 9n + 7 = 0\]

 

Answer 1

\[\text{ Given }: \]                                                                          \[(1 + x )^{2n} \]
\[\text{ Thus, we have: } \]
\[ T_2 = T_{1 + 1} \]
\[ = ^{2n}{}{C}_1 x^1 \]
\[ T_3 = T_{2 + 1} \]
\[ = ^{2n}{}{C}_2 x^2 \]
\[ T_4 = T_{3 + 1} \]
\[ = ^{2n}{}{C}_3 x^3 \]
\[\text{ We have coefficients of the 2nd, 3rd and 4th terms in AP } . \]
\[ \therefore 2\left( ^{2n}{}{C}_2 \right) = ^t{2n}{}{C}_1 + ^{2n}{}{C}_3 \]
\[ \Rightarrow 2 = \frac{^{2n}{}{C}_1}{^{2n}{}{C}_2} + \frac{^{2n}{}{C}_3}{^{2n}{}{C}_2} \]
\[ \Rightarrow 2 = \frac{2}{2n - 1} + \frac{2n - 2}{3}\]
\[ \Rightarrow 12n - 6 = 6 + 4 n^2 - 4n - 2n + 2\]
\[ \Rightarrow 4 n^2 - 18n + 14 = 0\]
\[ \Rightarrow 2 n^2 - 9n + 7 = 0\]