If frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} is | vedclass

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(C) Given that $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} = \frac{a + b}{2}$.Cross-multiplying,we get:$2(a^{n + 1} + b^{n + 1}) = (a + b)(a^n + b^n)$$2a

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(C) Given that $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} = \frac{a + b}{2}$.Cross-multiplying,we get:$2(a^{n + 1} + b^{n + 1}) = (a + b)(a^n + b^n)$$2a^{n + 1} + 2b^{n + 1} = a^{n + 1} + ab^n + ba^n + b^{n + 1}$Rearranging the terms:$a^{n + 1} - ab^n - ba^n + b^{n + 1} = 0$$a^n(a - b) - b^n(a - b) = 0$$(a^n - b^n)(a - b) = 0$Since $a 
eq b$,we must have $a^n - b^n = 0$,which implies $a^n = b^n$.Thus,$(a/b)^n = 1 = (a/b)^0$.Therefore,$n = 0$.

If $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$ is the $A.M.$ of $a$ and $b$,then $n = $

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