# How do you find the first 3 non-zero terms for the expansion of (9+x^2)^ (1/2)?

Apr 29, 2017

${\left(9 + {x}^{2}\right)}^{\frac{1}{2}} = 3 + {x}^{2} / 6 - {x}^{4} / 216 + {x}^{6} / 3888. \ldots \ldots \ldots \ldots . .$

#### Explanation:

Binomial expansion of ${\left(1 + x\right)}^{n} = {C}_{0}^{n} {a}^{n} + {C}_{1}^{n} x + {C}_{2}^{n} {x}^{2} + {C}_{3}^{n} {x}^{3} + . . + + {C}_{n}^{n} {x}^{n}$,

where ${C}_{r}^{n} = \frac{n \left(n - 1\right) \left(n - 2\right) \ldots . . \left(n - r + 1\right)}{1.2 .3 \ldots . . r}$ and ${C}_{0}^{n} = 1$

When we have $n$ as fraction this is written as infinite series.

Hence ${\left(9 + {x}^{2}\right)}^{\frac{1}{2}}$ can be written as

${9}^{\frac{1}{2}} {\left(1 + {x}^{2} / 9\right)}^{1} / 2$

= $3 \left[1 + \frac{\frac{1}{2}}{1} \left({x}^{2} / 9\right) + \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right)}{1 \cdot 2} {\left({x}^{2} / 9\right)}^{2} + \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right)}{1 \cdot 2 \cdot 3} {\left({x}^{2} / 9\right)}^{3} + \ldots \ldots \ldots \ldots\right]$

= $3 \left[1 + {x}^{2} / 18 + \frac{\frac{1}{2} \cdot \left(- \frac{1}{2}\right)}{2} \frac{{x}^{4}}{81} + \frac{\frac{1}{2} \cdot \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right)}{6} \frac{{x}^{6}}{729.} \ldots \ldots . .\right]$

= $3 \left[1 + {x}^{2} / 18 - {x}^{4} / 648 + {x}^{6} / 11664. \ldots \ldots \ldots \ldots . .\right]$

= $3 + {x}^{2} / 6 - {x}^{4} / 216 + {x}^{6} / 3888. \ldots \ldots \ldots \ldots . .$