If n is an even positive integer,then the con | vedclass

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Question

(A) If $n$ is even,the greatest coefficient is $^nC_{n/2}$.For the greatest term to have the greatest coefficient,the term $T_{n/2+1} = ^nC_{n/2} x^{n

Vedclass · Accepted Answer

$\frac{n}{n + 2} < x < \frac{n + 2}{n}$

Vedclass · Answer

(A) If $n$ is even,the greatest coefficient is $^nC_{n/2}$.For the greatest term to have the greatest coefficient,the term $T_{n/2+1} = ^nC_{n/2} x^{n/2}$ must be greater than its neighbors $T_{n/2}$ and $T_{n/2+2}$.$T_{n/2+1} > T_{n/2} \implies ^nC_{n/2} x^{n/2} > ^nC_{n/2-1} x^{n/2-1} \implies x > \frac{^nC_{n/2-1}}{^nC_{n/2}} = \frac{n/2}{n - n/2 + 1} = \frac{n/2}{n/2 + 1} = \frac{n}{n + 2}$.$T_{n/2+1} > T_{n/2+2} \implies ^nC_{n/2} x^{n/2} > ^nC_{n/2+1} x^{n/2+1} \implies x Thus,the condition is $\frac{n}{n + 2} < x < \frac{n + 2}{n}$.

If $n$ is an even positive integer,then the condition that the greatest term in the expansion of $(1 + x)^n$ may have the greatest coefficient also,is

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