Simplify \( \left(1 / 3^{x}-2\right)^{-1 / 2} \) using binomial theorem
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To simplify \( \left(1 / 3^{x}-2\right)^{-1 / 2} \) using the binomial theorem, we can rewrite the expression in a more convenient form. We have: \[ \left(1 - 6 \cdot 3^{-x}\right)^{-1/2} \] This is because \( 1 / 3^{x} - 2 \) can be rewritten as \( (1 - 6 \cdot 3^{-x}) \) after combining terms. Now, we can use the binomial theorem which states that for any real number \( n \) and \( |y| < 1 \): \[ (1 + y)^n \approx 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \ldots \] Applying this to our case with \( n = -\frac{1}{2} \) and \( y = -6 \cdot 3^{-x} \), we get: \[ \left(1 - 6 \cdot 3^{-x}\right)^{-1/2} \approx 1 + \left(-\frac{1}{2}\right)(-6 \cdot 3^{-x}) + \frac{-\frac{1}{2} \cdot (-\frac{3}{2})}{2!}(-6 \cdot 3^{-x})^2 + \ldots \] Calculating the first few terms gives us: 1. The first term: \( 1 \) 2. The second term: \( 3 \cdot 3^{-x} = 3^{1-x} \) 3. The third term: \[ \frac{-\frac{1}{2} \cdot (-\frac{3}{2})}{2}(36 \cdot 3^{-2x}) = \frac{9}{4} \cdot 36 \cdot 3^{-2x} = \frac{81}{4} \cdot 3^{-2x} \] Our approximation becomes: \[ \left(1 - 6 \cdot 3^{-x}\right)^{-1/2} \approx 1 + 3^{1-x} + \frac{81}{4} \cdot 3^{-2x} + \ldots \] Thus, \[ \left(1 / 3^{x}-2\right)^{-1 / 2} \approx 1 + 3^{1-x} + \frac{81}{4} \cdot 3^{-2x} + \ldots \] This provides a simplified series expansion using the binomial theorem!
