Four algorithms to find the Nth Fibonacci number

I'm implementing some basic algorithms in Go as an introductory exercise. Here are four different algorithms to find the Nth Fibonacci number.

I'm looking for general feedback, but I'm specially interested in the Go idiom. Are the algorithms implemented idiomatically? If not, how can I correctly use the Go idiom to implement them?

Any other feedback you can think of is also welcome.

// Algorithms to calculate the nth fibonacci number

package main

import (
    "fmt"
    "math"
)

func main() {
    for i := 0; i <= 20; i++ {
        fmt.Println(fibonacci(i))
        fmt.Println(fibonacciRecursive(i))
        fmt.Println(fibonacciTail(i))
        fmt.Println(fibonacciBinet(i))
        println()
    }
}

// Iterative
func fibonacci(n int) int {
    current, prev := 0, 1
    for i := 0; i < n; i++ {
        current, prev = current + prev, current
    }
    return current
}

// Recursive
func fibonacciRecursive(n int) int {
    if n < 2 {
        return n
    } 
    return fibonacciRecursive(n - 1) + fibonacciRecursive(n - 2)
}

// Tail recursion
func fibonacciTail(n int) int {
    return tailHelper(n, 1, 0)
}
func tailHelper(term, val, prev int) int {
    if term == 0 {
        return prev
    }
    if term == 1 {
        return val
    }
    return tailHelper(term - 1, val + prev, val)
}

// Analytic (Binet's formula)
func fibonacciBinet(num int) int {
    var n float64 = float64(num);
    return int( ((math.Pow(((1 + math.Sqrt(5)) / 2), n) - math.Pow(1 - ((1 + math.Sqrt(5)) / 2), n)) / math.Sqrt(5)) + 0.5 )
}

For tailHelper(), the term == 1 case is superfluous and should be eliminated.

In fibonacciBinet(), the quantity \$\frac{1 + \sqrt{5}}{2}\$ appears twice. Since that quantity is known as the Golden Ratio, I would define an intermediate value

var phi = (1 + math.Sqrt(5)) / 2;

However, none of these four solutions is particularly idiomatic for Go. The most expressive way to write a Fibonacci sequence in Go is as an iterator, such as this one.

StackExchange Code Review Q#48020, answer score: 4