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simply 3^n+3 + 3 ^n / 3^n+3 - 3^n+1 |
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Answer» \(\frac{3^{n+3}+3^n}{3^{n+3}-3^{n+1}}\) \(=\frac{3^n(3^3+1)}{3^{n+1}{(3^2-1)}}\) \(=\frac{1}{3}\times\frac{27+1}{9-1}\) \(=\frac{28}{3\times8}\) \(=\frac{7}{6}\) |
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