Using calculus, we can show that $$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } + \ldots , \text { for } | x |

Question

Using calculus, we can show that $$( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } + \ldots , \text { for } | x | < 1$$ This
formula can be used to evaluate binomial expressions raised to noninteger exponents. Use the first four
terms of this infinite series to approximate the given expression. Round to 3 decimal places if necessary.
-$$( 1.3 ) ^ { 4 / 3 }$$
A) 1.423
B) 1.416
C) 1.419
D) 1.422

Quizplus · Accepted Answer

The given expression can be approximated using the binomial series expansion formula with $n = \frac{4}{3}$ and $x = 0.3$. Substituting these values into the first four terms of the series gives: $$1 + \frac{4}{3}(0.3) + \frac{\frac{4}{3}(\frac{4}{3}-1)}{2!}(0.3)^2 + \frac{\frac{4}{3}(\frac{4}{3}-1)(\frac{4}{3}-2)}{3!}(0.3)^3$$ Calculating this expression yields approximately 1.419.

Using Calculus, We Can Show That This Formula Can Be Used to Evaluate Binomial Expressions Raised

Using Calculus, We Can Show That This
Formula Can Be Used to Evaluate Binomial Expressions Raised