Question

If the coefficient of $2$nd, $3$rd and $4$th terms in the expansion of ${(1+x)}^{2n}$ are in A.P. Show that $2n^2-9n+7=0$

Answer 1

$(1+x{)}^{2n}$

$2ndterm=\left(2n\right)c_{1}x$

$3rdterm=\left(2n\right)c_2{x}^2$

$4thterm=\left(2n\right)c_3{x}^2$

coefficient in AP $\Rightarrow \left(2n\right)c_1+\left(2n\right)c^3=2\times \left(2n\right)c_2$

$\therefore 2n+\frac{\left(2n\right)(2n-1)(2n-2)}{3!}=2\times \frac{2n(2n-1)}{2!}$

$\therefore 2n(1+\frac{(2n-1)(2n-2)}{6})=2n(2n-1)$

$\therefore (6+4n^{2-6n+2})=12n-6$

$\therefore 4n^2-18n+14=0$

$\therefore 2n^2-9n+7=0$