How do you use the binomial series to expand #f(x) = (6-x)^-3#?

1 Answer
Apr 17, 2018

Answer:

#1/216-1/432x+1/1296x^2-5/23328x^3+5/93312x^4-7/559872x^5+...#

Explanation:

The binomial expansion when #n# is fractional or negative is a little different in form than when #n# is positive. We are not able to use the standard method of #color(white)(0)^nC_r#, since this is only valid for positive integers. Notice also that we do not have a finite number of terms. The number of terms required would have to be specified.

The form we will use is:

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

We have #(6+x)^-3#, we need to manipulate this to get the required form:

#1/6^3(1+x/6)^-3#

#1+(-3)(x/6)+((-3)(-4))/2(x/6)^2+((-3)(-4)(-5))/6(x/6)^3->#

#+((-3)(-4)(-5)(-6))/24(x/6)^4+((-3)(-4)(-5)(-6)(-7))/120(x/6)^5#

#1/216(1-1/2x+1/6x^2-5/108x^3+5/432x^4-7/2592x^5+...)#

#1/216-1/432x+1/1296x^2-5/23328x^3+5/93312x^4-7/559872x^5+...#

These series are always infinite, unless #|x|<1#, or in this case #|x/6|<1#