Discussion
If \(\frac{4^{n+1} + 16^{n+1}}{4^{n} + 16^{n}}\) is the Geometric Mean between 4 and 16, then the value of \(n\) is
Answer: D
\(n = \frac{-1}{2}\).
The geometric mean of 4 and 16 is \(\sqrt{4 \times 16} = 8\). Write \(a = 4^{n}\), so that \(16^{n} = a^{2}\):
\[ \begin{aligned} \frac{4a + 16a^{2}}{a + a^{2}} &= 8 \\ 4 + 16a &= 8 + 8a \quad (a \gt 0) \\ a &= \frac{1}{2} \end{aligned} \]
Then \(4^{n} = \frac{1}{2}\), so \(2^{2n} = 2^{-1}\) and \(n = -\frac{1}{2}\).
Check: \(4^{1/2} + 16^{1/2} = 6\) and \(4^{-1/2} + 16^{-1/2} = \frac{3}{4}\), and \(6 \div \frac{3}{4} = 8\). The option \(\frac{1}{2}\) has the right size but the wrong sign; it gives \(\frac{8 + 64}{2 + 4} = 12\).
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