Q58 / 160 · AP EAMCET 2016 · Mathematics

If the middle term in the expansion of $(1+x)^{2 n}$ is the greatest term, then $x$ lies in the interval

  • Correct answer
    A
    $\left(\frac{n}{n+1}, \frac{n+1}{n}\right)$
  • B
    $\left(\frac{n+1}{n}, \frac{n}{n+1}\right)$
  • C
    $(n-2, n)$
  • D
    $(n-1, n)$

Solution

In the expansion of $(1+x)^{2 n}$, middle term is ${ }^{2 n} \mathrm{C}_n x^n$. Since, middle term is the greatest term. $\therefore{ }^{2 n} \mathrm{C}_n x^n>{ }^{2 n} \mathrm{C}_{n-1} x^{n-1}$ $\text { and }{ }^{2 n} \mathrm{C}_r x^n>{ }^{2 n} \mathrm{C}_{n+1} x^{n+1}$ $\begin{aligned} & \Rightarrow \quad x>\frac{{ }^{2 n} \mathrm{C}_{n-1}}{{ }^{2 n} \mathrm{C}_n} \text { and } x < \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}} \\ & \text { Hence, } x \in\left(\frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}, \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}\right) \end{aligned}$ Hence, $x \in\left(\frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}, \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}\right)$ $\begin{aligned} & x \in\left(\begin{array}{c} \frac{(2 n) !}{(n-1) !(2 n-n+1)} \times \frac{n !(2 n-n) !}{(2 n) !} \\ \frac{(2 n) !}{n !(2 n-n) !} \times \frac{(n-1) !(2 n-n+1)}{(2 n) !} \end{array}\right) \\ & x \in\left(\frac{n(n-1) ! n !}{n(n-1) !(n+1) n !}, \frac{(n+1) n !(n-1) !}{n(n-1) ! n !}\right) \\ & \therefore \quad x \in\left(\frac{n}{n+1}, \frac{n+1}{n}\right) \end{aligned}$

Hence, the correct answer is option: 'A'.

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