How do you use the binomial series to expand # 1/((2+x)^3)#?

1 Answer
Feb 13, 2018

Answer:

The binomial expansion is
#= 1/8 (1- 3/2 x+3/2 x^2-5/4 x^3+...)# for #|x|<2# and

#= 1/x^3 (1-6/x+24/x^2- 80/x^3+...)# for #|x|>2#

Explanation:

One important point to remember is that if the index #n# is not a positive integer, the binomial expansion leads to an infinite series. In such a situation, it is important to keep in mind that the infinite binomial series for #(1+x)^n# converges only if #|x|<1#.

For an arbitrary #n#, the binomial expansion for #(1+x)^n# is given by

#1+n x+ {n(n-1)}/{2!} x^2+ {n(n-1)(n-2)}/{3!} x^3 +... #

So we first recast the expression that we want to expand binomially in the form #(1+x)^n#

Here we first use
#1/{(2+x)^3} = 1/2^3 (1+x/2)^{-3} #

and then use the binomial expansion for #n=-3# to get

#1/{(2+x)^3} = 1/2^3 (1+(-3) x/2 + {(-3)(-4)}/{2!} (x/2)^2 + {(-3)(-4)(-5)}/{3!} (x/2)^3 +.... )#
#= 1/8 (1- 3/2 x+3/2 x^2-5/4 x^3+...)#

Of course this infinite series only converges for #|x/2|<1#, i. e. #-2 < x <2#.

If #|x/2|>1#, this series will diverge - rendering it useless! What saves the day is that in this case, #|2/x|<1# and so, we can still get a binomial expansion, but thus time we must start from
#1/{(2+x)^3} = 1/x^3 (1+2/x)^{-3} #
and continue with

#1/{(2+x)^3} = 1/x^3 (1+(-3) 2/x + {(-3)(-4)}/{2!} (2/x)^2 + {(-3)(-4)(-5)}/{3!} (2/x)^3 +.... )#
#= 1/x^3 (1-6/x+24/x^2-80/x^3+...)#