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

Essay

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.
- $$4 + 4 \cdot \frac { 1 } { 7 } + 4 \cdot \left( \frac { 1 } { 7 } \right) ^ { 2 } + \ldots + 4 \cdot \left( \frac { 1 } { 7 } \right) ^ { n - 1 } = \frac { 4 \left( 1 - \left( \frac { 1 } { 7 } \right) ^ { n } \right) } { 1 - \frac { 1 } { 7 } }$$