How do you use the binomial series to expand #g (x) = 8 / (1 - x) ^3#?

1 Answer
Nov 11, 2016

Answer:

#g(x)=8+24x+48x^2+80x^3+120x^4+168x^5+224x^6+....#

Explanation:

According to binomial series #(1+a)^n=1+na+(n(n-1))/(2!)a^2+(n(n-1)(n-2))/(3!)a^3+(n(n-1)(n-2)(n-3))/(4!)a^4+.......#

Hence #g(x)=8/(1-x)^3=8xx(1-x)^(-3)#

Now #(1-x)^(-3)=(1+(-x))^(-3)=1+(-3)(-x)+((-3)(-4))/1.2(-x)^2+((-3)(-4)(-5))/(1.2.3)(-x)^3+((-3)(-4)(-5)(-6))/(1.2.3.4)(-x)^4+((-3)(-4)(-5)(-6)(-7))/(1.2.3.4.5)(-x)^5+((-3)(-4)(-5)(-6)(-7)(-8))/(1.2.3.4.5.6)(-x)^6+.......#

or #=1+3x+6x^2+10x^3+15x^4+21x^5+28x^6+.......#

Hence #g(x)=8/(1-x)^3=8xx(1+3x+6x^2+10x^3+15x^4+21x^5+28x^6+.......)#

= #8+24x+48x^2+80x^3+120x^4+168x^5+224x^6+.......#