Binomially expand the equation (2+kx)^-3
(2+kx)-3 = (2-3)(1+kx/2)-3 = (2-3)(1+(-3)(kx/2) + [(-3)(-4)]/2! (kx/2)2 + [(-3)(-4)(-5)]/3! (kx/2)3 +... )
= 1⁄8 [1 -(3kx/2) + (12⁄2 k2x2/4) + (60⁄6 k3x3/8) + ...]
= 1⁄8 [1 - (3⁄2 kx) + ( 3⁄2 k2x2) + (5⁄4 k3x3) + ...]
= 1⁄8 - 3⁄16 kx + 3⁄16 k2x2 + 5⁄32 k3x3
CH
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