Use the binomial series to find the expansion of 1/(2+5x)^3 in ascending powers of x up to x^3 (|x|<2/5)
We want to rearrange the expression to the form (1+y)^n so we can use the general result: (1+y)^n=1+ny+[n(n-1)/2]y^2+[n(n-1)(n-2)/3!]y^3+... 1/(2+5x)^3 = (2+5x)^-3 = [2(1+5x/2)]^-3 = (2^-3)(1+5x/2)^-3 using the result ... = (1/8)(1+(-3)(5x/2)+(-3)(-4)/2^2+(-3)(-4)(-5)/3!^3+... = (1/8)(1-15x/2+(75/2)x^2-(625/4)x^3)= 1/8-(15/16)x+(75/16)x^2-(625/32)x^3
SJ
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