The value of n for which frac{x^{n+1}+y^{n+1} | vedclass

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Question

(A) Given that $\frac{x^{n+1}+y^{n+1}}{x^n+y^n} = \sqrt{xy}$.This implies $x^{n+1} + y^{n+1} = (xy)^{1/2} (x^n + y^n)$.$x^{n+1} + y^{n+1} = x^{n+1/2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given that $\frac{x^{n+1}+y^{n+1}}{x^n+y^n} = \sqrt{xy}$.This implies $x^{n+1} + y^{n+1} = (xy)^{1/2} (x^n + y^n)$.$x^{n+1} + y^{n+1} = x^{n+1/2} y^{1/2} + x^{1/2} y^{n+1/2}$.Rearranging the terms,we get $x^{n+1} - x^{n+1/2} y^{1/2} = x^{1/2} y^{n+1/2} - y^{n+1}$.$x^{n+1/2} (x^{1/2} - y^{1/2}) = y^{n+1/2} (x^{1/2} - y^{1/2})$.Assuming $x 
eq y$,we can divide both sides by $(x^{1/2} - y^{1/2})$:$x^{n+1/2} = y^{n+1/2}$.$(x/y)^{n+1/2} = 1$.Since $(x/y)^0 = 1$,we have $n + 1/2 = 0$.Therefore,$n = -1/2$.

The value of $n$ for which $\frac{x^{n+1}+y^{n+1}}{x^{n}+y^{n}}$ is the geometric mean of $x$ and $y$ is

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