(b) ${T_2} = {}^{2n}{C_1}\,\,x$, ${T_3} = {}^{2n}{C_2}\,\,{x^2}$, ${T_4} = {}^{2n}{C_3}\,\,{x^3}$

Coefficient of $T_2, T_3, T_4$ are in $A.P$.

==> $2.{}^{2n}{C_2} = {}^{2n}{C_1} + {}^{2n}{C_3}$

==> $2\frac{{2n!}}{{2\,!\,(2n – 2)\,!}} = \frac{{2n!}}{{(2n – 1)\,!}} + \frac{{2n!}}{{3\,!\,(2n – 3)\,!}}$

==> $\frac{{2\,.\,2n(2n – 1)}}{2} = 2n + \frac{{\,2n(2n – 1)(2n – 2)}}{6}$

==> $n(2n – 1) = n + \frac{{(n)(2n – 1)(2n – 2)}}{6}$

==> $6(2{n^2} – n) = 6n + 4{n^3} – 6{n^2} + 2n$

==> $6n(2n – 1) = 2n(2{n^2} – 3n + 4)$

==> $6n – 3 = 2{n^2} – 3n + 4$

==> $0 = 2{n^2} – 9n + 7$

==> $2{n^2} – 9n + 7 = 0$.