Display powers of Phi with Fibonacci precision
dc, 26 bytes

99k5v1+2/?^d5v/.5+0k1/k1/p


Due to the initial precision of 99 digits after the comma, this will work up yo input 11. A dynamic (or higher static) precision is possible, but would elevate the byte count.

Test cases

$ for ((i = 0; i &lt; 11; i++)) { dc -e '99k5v1+2/?^d5v/.5+0k1/k1/p' &lt;&lt;&lt; $i; }
1
1.6
2.6
4.23
6.854
11.09016
17.94427190
29.0344418537486
46.978713763747791812296
76.0131556174964248389559523684316960
122.9918693812442166512522758901100964746170048893169574174


How it works

Since the desired output is &phi;n, we can calculate the the Fibonacci number F(n) as &lfloor;&phi;n ÷ &radic;5 + 0.5&rfloor; with little additional effort.

99k                         Set the precision to 99.
   5v                       Compute the square root of 5.
     1+                     Add 1.
       2/                   Divide by 2.
                            This pushes the golden ratio.
         ?                  Read the input from STDIN.
          ^                 Elevate the golden ratio to that power.
           d                Push a copy.
            5v/             Divide it by the square root of 5.
               .5+          Add 0.5.
                  0k        Set the precision to 0.
                    1/      Divide by 1, truncating to the desired precision.
                            This pushes F(n).
                      k     Set the precision to F(n).
                       1/   Divide by 1, truncating to the desired precision.
                         p  Print.