If the coefficients of 2nd, 3rd and 4th terms in the expansion

Solution:

Given:

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

Thus, we have:

$T_{2}=T_{1+1}$

$={ }^{2 n} C_{1} x^{1}$

$T_{3}=T_{2+1}$

$={ }^{2 n} C_{2} x^{2}$

$T_{4}=T_{3+1}$

$={ }^{2 n} C_{3} x^{3}$

We have coefficients of the 2 nd, 3 rd and 4 th terms in AP.

$\therefore 2\left({ }^{2 n} C_{2}\right)={ }^{2 n} C_{1}+{ }^{2 n} C_{3}$

$\Rightarrow 2=\frac{{ }^{2 n} C_{1}}{{ }^{2 n} C_{2}}+\frac{{ }^{2 n} C_{3}}{{ }^{2 n} C_{2}}$

$\Rightarrow 2=\frac{2}{2 n-1}+\frac{2 n-2}{3}$

$\Rightarrow 12 n-6=6+4 n^{2}-4 n-2 n+2$

$\Rightarrow 4 n^{2}-18 n+14=0$

$\Rightarrow 2 n^{2}-9 n+7=0$

Hence proved.