If the middle term in the expansion of (1+x)^ | vedclass

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(A) In the expansion of $(1+x)^{2n}$,the middle term is $T_{n+1} = {}^{2n}C_n x^n$. Since the middle term is the greatest term,it must be greater than

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$\left(\frac{n}{n+1}, \frac{n+1}{n}\right)$

Vedclass · Answer

(A) In the expansion of $(1+x)^{2n}$,the middle term is $T_{n+1} = {}^{2n}C_n x^n$. Since the middle term is the greatest term,it must be greater than or equal to its adjacent terms $T_n$ and $T_{n+2}$.$T_{n+1} \ge T_n \implies {}^{2n}C_n x^n \ge {}^{2n}C_{n-1} x^{n-1} \implies x \ge \frac{{}^{2n}C_{n-1}}{{}^{2n}C_n} = \frac{n}{n+1}$.$T_{n+1} \ge T_{n+2} \implies {}^{2n}C_n x^n \ge {}^{2n}C_{n+1} x^{n+1} \implies x \le \frac{{}^{2n}C_n}{{}^{2n}C_{n+1}} = \frac{n+1}{n}$.Thus,$x \in \left[\frac{n}{n+1}, \frac{n+1}{n}\right]$. Given the options,the interval is $\left(\frac{n}{n+1}, \frac{n+1}{n}\right)$.

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

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