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

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 \( |x| < 1 \).