# How do you use the binomial series to expand (1+x)^(3/2)?

Sep 16, 2017

${\left(1 + x\right)}^{\frac{3}{2}} = 1 + \frac{3}{2} x + \frac{3}{8} {x}^{2} - \frac{1}{16} {x}^{3} + \frac{3}{128} {x}^{4} + \ldots$

#### Explanation:

The binomial series tell us that:

 (1+x)^n = 1+nx + (n(n-1))/(2!)x^2 + (n(n-1)(n-2))/(3!)x^3 + ...

And so for the given function, Let:

$f \left(x\right) = {\left(1 + x\right)}^{\frac{3}{2}}$

Then with $n = \frac{3}{2}$, we have:

 f(x) = 1 + (3/2)x + ((3/2)(1/2))/(2!)x^2 +
 \ \ \ \ \ \ \ \ \ \ \ \ ((3/2)(1/2)(-1/2))/(3!)x^3 +
 \ \ \ \ \ \ \ \ \ \ \ \ ((3/2)(1/2)(-1/2)(-3/2))/(4!)x^4 + ...

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1 + \frac{3}{2} x + \frac{\frac{3}{4}}{2} {x}^{2} + \frac{- \frac{3}{8}}{6} {x}^{3} + \frac{\frac{9}{16}}{24} {x}^{4} + \ldots$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 1 + \frac{3}{2} x + \frac{3}{8} {x}^{2} - \frac{1}{16} {x}^{3} + \frac{3}{128} {x}^{4} + \ldots$