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 |

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.
-$( 1.03 ) ^ { - 3 }$

Quizplus · Accepted Answer

To approximate $(1.03)^{-3}$ using the given series, we set $x = 0.03$ and $n = -3$. The series becomes $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$. Substituting $n = -3$ and $x = 0.03$, the first few terms are $1 - 3(0.03) + \frac{-3(-4)}{2}(0.03)^2 - \frac{-3(-4)(-5)}{6}(0.03)^3 + \ldots$. Calculating these terms and summing them gives an approximation close to $0.915$, which is to the nearest thousandth.

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 $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots$