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.
- $( 1.03 ) ^ { - 5 }$

Question

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.
-$$( 1.03 ) ^ { - 5 }$$
A) 0.863
B) 1.159
C) 1.154
D) 0.867

Quizplus · Accepted Answer

The Answer of 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.
-$$( 1.03 ) ^ { - 5 }$$
A) 0.863
B) 1.159
C) 1.154
D) 0.867

It Can Be Shown That (1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3…( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots(1+x)n=1+nx+2!n(n−1)​x2+3!n(n−1)(n−2)​x3…

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