How do you find the Maclaurin series for #3/(1-2x) #?

1 Answer
Narad T.
Oct 28, 2016

The series is #3(1+2x+4x^2+8x^3+16x^4+......)#

Explanation:

Rewriting the expression
#3/(1-2x)=3(1-2x)^(-1)#
The development of a binomial is
#(1+x)^n=1+nx+(n(n-1))/(1.2)x^2+(n(n-1)(n-2))/(1.2.3)x^3+(n(n-1)(n-3)(n-4))/(1.2.3.4)x^2+.....#

So we have
#(1-2x)^(-1)=1+2x+(-1(-2))/(1.2)(-2x)^2+(-1(-2)(-3))/(1.2.3)(-2x)^3+(-1(-2)(-3)(-4))/(1.2.3.4)(-2x)^2+.....#

#=1+2x+4x^2+8x^3+16x^4+.....#

and finally
#3/(1-2x)=3(1-2x)^(-1)=3(1+2x+4x^2+8x^3+16x^4+....)#

Related questions
See all questions in Constructing a Maclaurin Series
Impact of this question
5710 views around the world
You can reuse this answer
Creative Commons License