Question

If the greatest term in the expansion of ${(1+x)}^{2n}$ has also the greatest coefficient then $x$ lies between

• A. $\frac{n}{n+1}$ and$\frac{n+1}{n-1}$.
• B. $\frac{n}{n+1}$ and $\frac{n+1}{n^2}$.
• C. $\frac{n}{n-1}$ and $\frac{n+1}{n}$.
• D. $\frac{n}{n+1}$ and $\frac{n+1}{n}$.

Answer 1

Answer: (D)

$\frac{n}{n+1}$ and $\frac{n+1}{n}$.

In the expansion of ${(1+x)}^{2n}$ , the middle term is ${(\frac{2n}{2}+1)}^{th}$ i.e., $(n+1)th$ term, we kow that from binomial expansion, middle term has greatest coeffecient. ${(\because TermsT}_1,T_2,T_3....T_n,T_{n+1},T_{n+2},....)$<br><br>$\therefore T_n<T_{n+1}>T_{n+2}{T}_n<T_{n+1}{T}_{n+1}>T_n\therefore \frac{T_{n+1}}{T_n}>1(\because \frac{T_{r+1}}{T_r}=\frac{n-r+1}{r}x,heren=2n)\Rightarrow \frac{2n-n+1}{n}x>1\therefore x>\frac{n}{n+1}...........\left(i\right)andif{T}_{n+1}>T_{n+2}\therefore \frac{T_{n+2}}{T_{n+1}}<1\Rightarrow \frac{2n-(n+1)+1}{n}x<1\therefore x<\frac{n+1}{n}$ .........(ii)