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.
- $$4 \cdot 6 + 5 \cdot 7 + 6 \cdot 8 + \ldots + ( n + 3 ) ( n + 5 ) = \frac { n \left( 2 n ^ { 2 } + 27 n + 115 \right) } { 6 }$$