How do you use the binomial series to expand g (x) = 8 / (1 - x) ^3?

Nov 11, 2016

$g \left(x\right) = 8 + 24 x + 48 {x}^{2} + 80 {x}^{3} + 120 {x}^{4} + 168 {x}^{5} + 224 {x}^{6} + \ldots .$

Explanation:

According to binomial series (1+a)^n=1+na+(n(n-1))/(2!)a^2+(n(n-1)(n-2))/(3!)a^3+(n(n-1)(n-2)(n-3))/(4!)a^4+.......

Hence $g \left(x\right) = \frac{8}{1 - x} ^ 3 = 8 \times {\left(1 - x\right)}^{- 3}$

Now ${\left(1 - x\right)}^{- 3} = {\left(1 + \left(- x\right)\right)}^{- 3} = 1 + \left(- 3\right) \left(- x\right) + \frac{\left(- 3\right) \left(- 4\right)}{1.2} {\left(- x\right)}^{2} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right)}{1.2 .3} {\left(- x\right)}^{3} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right) \left(- 6\right)}{1.2 .3 .4} {\left(- x\right)}^{4} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right) \left(- 6\right) \left(- 7\right)}{1.2 .3 .4 .5} {\left(- x\right)}^{5} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right) \left(- 6\right) \left(- 7\right) \left(- 8\right)}{1.2 .3 .4 .5 .6} {\left(- x\right)}^{6} + \ldots \ldots .$

or $= 1 + 3 x + 6 {x}^{2} + 10 {x}^{3} + 15 {x}^{4} + 21 {x}^{5} + 28 {x}^{6} + \ldots \ldots .$

Hence $g \left(x\right) = \frac{8}{1 - x} ^ 3 = 8 \times \left(1 + 3 x + 6 {x}^{2} + 10 {x}^{3} + 15 {x}^{4} + 21 {x}^{5} + 28 {x}^{6} + \ldots \ldots .\right)$

= $8 + 24 x + 48 {x}^{2} + 80 {x}^{3} + 120 {x}^{4} + 168 {x}^{5} + 224 {x}^{6} + \ldots \ldots .$