Question

If the coefficient of second, third and fourth terms in the expansion if $(1+x{)}^{2n}$ are in A.P, the $2n^2-3n$ is equal to

• A. $-7$
• B. $14$
• C. $6$
• D. $-6$

Answer 1

The correct option is B $14$

Coefficient of second, third and fourth terms in the expansion of $(1+x{)}^{2n}$ is $(\frac{2n}{1}),(\frac{2n}{2}),(\frac{2n}{3})$ Also they are in A.P

So $2(\frac{2n}{2})=(\frac{2n}{1})+(\frac{2n}{3})$

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

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

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

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

$2n-1=6⟹n=\frac{7}{2}$

$2n^2-3n=2(\frac{7}{2}{)}^2-3\times \frac{7}{2}=14$