Question

The coefficient of 2 and 3and 4 term in the expansion of (1+x)^2n are in ap show that 2n^2_9n+7=0

Answer 1

$hola there

By the given condition we have $2nC_1,2nC_2,2nC_3$ are in A.P

$\therefore 2\times 2nC_2=2nC_1+2nC_3$

$\Rightarrow 2.\large\frac{2n(2n-1)}{1.2}=\frac{2n}{1}+\frac{2n(2n-1)(2n-2)}{1.2.3}$

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

$\Rightarrow 6n-3=3n+2n^2-3n+1$

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

Hence proved.