Use binomial theorem to sadre \( (1-x)^{-1 / 3} \)

To expand \( (1 - x)^{-1/3} \) using the binomial theorem, we can use the generalized binomial series, which states that for any real number \( n \): \[ (1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \] where \( \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} \). In our case, we have \( n = -\frac{1}{3} \) and we want to expand \( (1 - x)^{-1/3} \). We can rewrite it as: \[ (1 - x)^{-1/3} = (1 + (-x))^{-1/3} \] Now, we can apply the binomial series: \[ (1 + (-x))^{-1/3} = \sum_{k=0}^{\infty} \binom{-\frac{1}{3}}{k} (-x)^k \] Next, we need to calculate the binomial coefficients \( \binom{-\frac{1}{3}}{k} \): \[ \binom{-\frac{1}{3}}{k} = \frac{-\frac{1}{3} \left(-\frac{1}{3}-1\right) \left(-\frac{1}{3}-2\right) \cdots \left(-\frac{1}{3}-(k-1)\right)}{k!} \] Now, let's compute the first few terms of the series expansion. The first few terms will give us a good approximation of the function. 1. For \( k = 0 \): \[ \binom{-\frac{1}{3}}{0} = 1 \] 2. For \( k = 1 \): \[ \binom{-\frac{1}{3}}{1} = -\frac{1}{3} \] 3. For \( k = 2 \): \[ \binom{-\frac{1}{3}}{2} = \frac{-\frac{1}{3} \cdot -\frac{4}{3}}{2!} = \frac{\frac{4}{9}}{2} = \frac{2}{9} \] 4. For \( k = 3 \): \[ \binom{-\frac{1}{3}}{3} = \frac{-\frac{1}{3} \cdot -\frac{4}{3} \cdot -\frac{7}{3}}{3!} = \frac{-\frac{28}{27}}{6} = -\frac{14}{81} \] Now we can write the first few terms of the expansion: \[ (1 - x)^{-1/3} \approx 1 - \frac{1}{3}(-x) + \frac{2}{9}(-x)^2 - \frac{14}{81}(-x)^3 + \cdots \] This simplifies to: \[ (1 - x)^{-1/3} \approx 1 + \frac{1}{3}x + \frac{2}{9}x^2 - \frac{14}{81}x^3 + \cdots \] Thus, the expansion of \( (1 - x)^{-1/3} \) using the binomial theorem is: \[ (1 - x)^{-1/3} = \sum_{k=0}^{\infty} \binom{-\frac{1}{3}}{k} (-x)^k \] with the first few terms being: \[ 1 + \frac{1}{3}x + \frac{2}{9}x^2 - \frac{14}{81}x^3 + \cdots \]