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Show that (1+x)^-1/2 = 1 -1/2 x + 1 middot 3/2 middot 4 X^2 - 1 middot 3 middot 5/2 middot 4...

Question:

Show that {eq}(1+x)^{-1/2} = 1 -\frac{1}{2 x} + \frac{1 \cdot3}{2 \cdot 4} X^2 - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} X^3 +... {/eq}

Binomial Theorem:

For {eq}|x|<1 \text{and non integer}\hspace{.3cm}n, {/eq}

The binomial expansion for {eq}(1+x)^n {/eq} is given by

{eq}(1+x)^n = 1+nx +\frac{n(n-1)}{2\!}x^2 +\frac{n(n-1)(n-2)}{3\!}x^3 +.... {/eq}

Answer and Explanation:

Applying the binomial theorem:

{eq}(1+x)^{-\frac{1}{2}} = 1-\frac{1}{2}x +\frac{(-\frac{1}{2})((-\frac{1}{2})-1)}{2\!}x^2 +\frac{(-\frac{1}{2})((-\frac{1}{2})-1)((-\frac{1}{2})-2)}{3\!}x^3 +.... \\ = 1-\frac{1}{2}x + \frac{1.3}{2.4} x^2 - \frac{1.3.5}{2.4.6} x^3 +... {/eq}

Hence proved


Learn more about this topic:

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The Binomial Theorem: Defining Expressions

from Algebra II: High School

Chapter 12 / Lesson 7
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