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The value of $\mathrm{n}$ for which $\frac{\mathrm{x}^{\mathrm{n}+1}+\mathrm{y}^{\mathrm{n}+1}}{\mathrm{x}^{\mathrm{n}}+\mathrm{y}^{\mathrm{n}}}$ is the geometric mean of $\mathrm{x}$ and $\mathrm{y}$ is
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Verified Answer
The correct answer is:
$\mathrm{n}=-\frac{1}{2}$
Hints: $\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=\sqrt{x y} \Rightarrow x^{n+1}+y^{n+1}=\sqrt{x y}\left(x^n+y^n\right)$
$$
\mathrm{x}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right)=\mathrm{y}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right),\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{n}+\frac{1}{2}}=1 \quad \mathrm{n}=-\frac{1}{2}
$$
$$
\mathrm{x}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right)=\mathrm{y}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right),\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{n}+\frac{1}{2}}=1 \quad \mathrm{n}=-\frac{1}{2}
$$
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