Find the value of n so that frac{a^{n+1}+b^{n | vedclass

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(A) The geometric mean $(GM)$ of $a$ and $b$ is $\sqrt{ab}$.By the given condition:$\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} = \sqrt{ab} = (ab)^{1/2}$Squar

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$n = -\frac{1}{2}$

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(A) The geometric mean $(GM)$ of $a$ and $b$ is $\sqrt{ab}$.By the given condition:$\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} = \sqrt{ab} = (ab)^{1/2}$Squaring both sides,we obtain:$\frac{(a^{n+1}+b^{n+1})^2}{(a^n+b^n)^2} = ab$$a^{2n+2} + 2a^{n+1}b^{n+1} + b^{2n+2} = ab(a^{2n} + 2a^nb^n + b^{2n})$$a^{2n+2} + 2a^{n+1}b^{n+1} + b^{2n+2} = a^{2n+1}b + 2a^{n+1}b^{n+1} + ab^{2n+1}$Subtracting $2a^{n+1}b^{n+1}$ from both sides:$a^{2n+2} + b^{2n+2} = a^{2n+1}b + ab^{2n+1}$Rearranging the terms:$a^{2n+2} - a^{2n+1}b = ab^{2n+1} - b^{2n+2}$$a^{2n+1}(a - b) = b^{2n+1}(a - b)$Assuming $a 
eq b$,we divide by $(a - b)$:$a^{2n+1} = b^{2n+1}$$(\frac{a}{b})^{2n+1} = 1 = (\frac{a}{b})^0$$2n + 1 = 0$$n = -\frac{1}{2}$

Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b$.

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