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If the coefficients of second, third and fourth term in the expansion of ${(1 + x)^{2n}}$ are in $A.P.$, then $2{n^2} - 9n + 7$ is equal to
$-1$
$0$
$1$
$3\over2$
Solution
(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 $A.P$.
==> $2.{}^{2n}{C_2} = {}^{2n}{C_1} + {}^{2n}{C_3}$
==> $2\frac{{2n!}}{{2\,!\,(2n – 2)\,!}} = \frac{{2n!}}{{(2n – 1)\,!}} + \frac{{2n!}}{{3\,!\,(2n – 3)\,!}}$
==> $\frac{{2\,.\,2n(2n – 1)}}{2} = 2n + \frac{{\,2n(2n – 1)(2n – 2)}}{6}$
==> $n(2n – 1) = n + \frac{{(n)(2n – 1)(2n – 2)}}{6}$
==> $6(2{n^2} – n) = 6n + 4{n^3} – 6{n^2} + 2n$
==> $6n(2n – 1) = 2n(2{n^2} – 3n + 4)$
==> $6n – 3 = 2{n^2} – 3n + 4$
==> $0 = 2{n^2} – 9n + 7$
==> $2{n^2} – 9n + 7 = 0$.
