It Can Be Shown That $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots$

Multiple Choice

It can be shown that $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots$ is true for any real number $n$ (not just positive integer values) and any real number $x$ , where $| x | < 1$ . Use this series to approximate the given number to the nearest thousandth.
- $$1.06 ^ { 3 }$$

A) 0.837
B) 1.191
C) 0.840
D) 1.194