Question

The coefficient of $2^{nd},3^{rd},4^{th}$ terms in the expansion of $(1+x{)}^{2n}$ are $A.P$., then $2n^2-9n+8=$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $1$

Given:- expansion-

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

Coefficient of $2^{nd}$ term $={}^{2n}{C}_1$

Coefficient of $3^{rd}$ term $={}^{2n}{C}_2$

Coefficient of $4^{th}$ term $={}^{2n}{C}_3$

Given that the coefficient of $2^{nd},3^{rd}$ and $4^{th}$ term are in A.P., thus

$2\times {}^{2n}{C}_2={}^{2n}{C}_1+{}^{2n}{C}_3$

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

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

$6(2n-1)=6+(4n^2-6n+2)$

$12n-6=4n^2-6n+8$

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

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

$\Rightarrow 2n^2-9n+7+1-1=0$

$\Rightarrow 2n^2-9n+8=1$