Use pascaf's triangle to solve $$ (1-x)^{-1 / 3} $$

Question

UpStudy AI · Accepted Answer

To solve $$ (1-x)^{-1/3} $$ using Pascal's triangle, we can use the binomial series expansion. The binomial series for $$ (1-x)^n $$ is given by:

$$
(1-x)^n = \sum_{k=0}^{\infty} \binom{n}{k} (-x)^k
$$

where $$ \binom{n}{k} $$ is the binomial coefficient, defined as:

$$
\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}
$$

For our case, we have $$ n = -\frac{1}{3} $$. Thus, we can write:

$$
(1-x)^{-1/3} = \sum_{k=0}^{\infty} \binom{-\frac{1}{3}}{k} (-x)^k
$$

Now, we need to calculate the first few terms of this series. The binomial coefficient for negative $$ n $$ can be calculated as follows:

$$
\binom{-\frac{1}{3}}{k} = \frac{(-\frac{1}{3})(-\frac{1}{3}-1)(-\frac{1}{3}-2)\cdots(-\frac{1}{3}-(k-1))}{k!}
$$

Let's calculate the first few terms of the series expansion for $$ k = 0, 1, 2, 3 $$:

1. For $$ k = 0 $$:
   $$
   \binom{-\frac{1}{3}}{0} = 1
   $$

2. For $$ k = 1 $$:
   $$
   \binom{-\frac{1}{3}}{1} = -\frac{1}{3}
   $$

3. For $$ k = 2 $$:
   $$
   \binom{-\frac{1}{3}}{2} = \frac{(-\frac{1}{3})(-\frac{4}{3})}{2!} = \frac{\frac{4}{9}}{2} = \frac{2}{9}
   $$

4. For $$ k = 3 $$:
   $$
   \binom{-\frac{1}{3}}{3} = \frac{(-\frac{1}{3})(-\frac{4}{3})(-\frac{7}{3})}{3!} = \frac{-\frac{28}{81}}{6} = -\frac{14}{243}
   $$

Now, substituting these values into the series expansion, we get:

$$
(1-x)^{-1/3} \approx 1 - \frac{1}{3}x + \frac{2}{9}x^2 - \frac{14}{243}x^3 + \cdots
$$

Thus, the expansion of $$ (1-x)^{-1/3} $$ using Pascal's triangle (binomial series) is:

$$
(1-x)^{-1/3} = 1 - \frac{1}{3}x + \frac{2}{9}x^2 - \frac{14}{243}x^3 + \cdots
$$

This series converges for $$ |x| < 1 $$.
$$ (1-x)^{-1/3} = 1 - \frac{1}{3}x + \frac{2}{9}x^2 - \frac{14}{243}x^3 + \cdots $$

Use pascaf's triangle to solve \( (1-x)^{-1 / 3} \)

Solución de inteligencia artificial de Upstudy

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Pre Calculus Questions