Question

If the coefficients 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

Given the expression $(1+x{)}^{2n}$ According to the condition

${}^{2n}{C}_1{,}^{2n}{C}_2{,}^{2n}{C}_3$ are in $A.P$

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

using ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

solving,

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

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

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

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

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

Hence, proved