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If $n$ is even posirive 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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Verified Answer
The correct answer is:
$\frac{n}{n+2} < x < \frac{n+2}{n}$
For greatest term of $(1+x)^{n},$ we have
$\quad \frac{n}{2} < \frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad \frac{n}{2} < \frac{n+1}{1+x}$ and $\frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad 1+x < \frac{n+1}{n / 2}$ and $\frac{n+1}{\frac{n}{2}+1} < 1+x$
$\Rightarrow \quad x < \frac{n+1-n / 2}{n / 2}$
and $\quad \frac{n+1-(n / 2-1)}{\frac{n+2}{2}} < x$
$\Rightarrow \quad x < \frac{n+2}{n}$ and $\frac{n}{n+2} < x$
$\Rightarrow \quad \frac{n}{n+2} < x < \frac{n+2}{n}$
$\quad \frac{n}{2} < \frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad \frac{n}{2} < \frac{n+1}{1+x}$ and $\frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad 1+x < \frac{n+1}{n / 2}$ and $\frac{n+1}{\frac{n}{2}+1} < 1+x$
$\Rightarrow \quad x < \frac{n+1-n / 2}{n / 2}$
and $\quad \frac{n+1-(n / 2-1)}{\frac{n+2}{2}} < x$
$\Rightarrow \quad x < \frac{n+2}{n}$ and $\frac{n}{n+2} < x$
$\Rightarrow \quad \frac{n}{n+2} < x < \frac{n+2}{n}$
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