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

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Question

(A) For the expansion $(1+x)^n$, the greatest coefficient occurs at the middle term$(s)$. Since $n$ is even, the greatest coefficient is at the term $

Vedclass · Accepted Answer

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

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(A) For the expansion $(1+x)^n$, the greatest coefficient occurs at the middle term$(s)$. Since $n$ is even, the greatest coefficient is at the term $T_{n/2+1}$.For the greatest term $T_{r+1}$ in the expansion of $(1+x)^n$, we have the condition $\frac{n-r+1}{r} x \ge 1$ and $\frac{n-r+1}{r+1} x \le 1$.For the greatest term to be the one with the greatest coefficient, we set $r = n/2$.Substituting $r = n/2$ into the inequality $\frac{n-r+1}{r} x > 1$ and $\frac{n-r+1}{r+1} x $\frac{n - n/2 + 1}{n/2} x > 1$ $\Rightarrow \frac{n/2 + 1}{n/2} x > 1$ $\Rightarrow \frac{n+2}{n} x > 1$ $\Rightarrow x > \frac{n}{n+2}$.$\frac{n - n/2 + 1}{n/2 + 1} 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 $(a+x)^{n}$ may also have the greatest coefficient is

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