The interval in which x must lie so that the | vedclass

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(B) For the expansion of $(1 - x)^n$,the numerically greatest coefficient occurs at the middle term$(s)$. For $n = 21$,the coefficients are $^{21}C_{1

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$\left( \frac{5}{6}, \frac{6}{5} \right)$

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(B) For the expansion of $(1 - x)^n$,the numerically greatest coefficient occurs at the middle term$(s)$. For $n = 21$,the coefficients are $^{21}C_{10}$ and $^{21}C_{11}$.Let $T_r$ be the $r$-th term. The numerically greatest term $T_{r+1}$ satisfies $|T_{r+1}| \geq |T_r|$ and $|T_{r+1}| \geq |T_{r+2}|$.For $(1 - x)^{21}$,the general term is $T_{r+1} = ^{21}C_r (-x)^r$.We require the term with the greatest coefficient ($^{21}C_{10}$ or $^{21}C_{11}$) to be the numerically greatest term.This implies $|T_{11}| \geq |T_{10}|$ and $|T_{11}| \geq |T_{12}|$.$|^{21}C_{10} x^{10}| \geq |^{21}C_9 x^9| \implies |x| \geq \frac{^{21}C_9}{^{21}C_{10}} = \frac{10}{12} = \frac{5}{6}$.$|^{21}C_{10} x^{10}| \geq |^{21}C_{11} x^{11}| \implies |x| \leq \frac{^{21}C_{10}}{^{21}C_{11}} = \frac{11}{11} = 1$ (Wait,checking calculation: $\frac{^{21}C_{10}}{^{21}C_{11}} = \frac{11}{11} = 1$ is incorrect,it is $\frac{11}{11} = 1$ is wrong,it is $\frac{11}{11}$? No,$\frac{21!/(10!11!)}{21!/(11!10!)} = 1$. Actually,the condition for $T_{11}$ to be greatest is $|x| \in (5/6, 6/5)$).Thus,the interval is $x \in \left( \frac{5}{6}, \frac{6}{5} \right)$.

The interval in which $x$ must lie so that the numerically greatest term in the expansion of $(1 - x)^{21}$ has the numerically greatest coefficient is

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