How can I write Mathematica code for this continued fraction with alternating terms?

You can use ContinuedFractionK[] for this:

φ + ContinuedFractionK[1, φ^(1 - 2 Boole[Mod[k, 2] == 1]), {k, n}]

Greg Martin suggests the simpler expression

φ + ContinuedFractionK[1, φ^((-1)^k), {k, n}]

An exercise for the motivated reader is to prove that this is equivalent to the simpler

(Fibonacci[n + 2] φ)/Fibonacci[n + 1]

z = Defer /@ {-1, ""};
φ + Nest[1/(φ^Last[z = RotateLeft[z]] + #) &, …, 5]

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z = Defer /@ {Style[-1, 14], ""};
Style[φ + Nest[1/(φ^Last[z = RotateLeft@z] + #) &, …, 5], 32, ScriptSizeMultipliers -> 1]

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Alternatively,

z = φ^(Defer /@ {Style[-1, 14], ""});
i = 1;
Style[φ + Nest[1/(z[[Mod[i++, 2, 1]]] + #) &, …, 5], 32, ScriptSizeMultipliers -> 1]

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You can use Fold[] for this (an example from Documentation Center):

ϕ + Fold[1/(#2 + #1) &, ϕ, Reverse[Table[ϕ^(1 - 2 Boole[Mod[k, 2] == 1]), {k, 1, 7}]]]

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