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

1 Answer
May 1, 2018

Answer:

#(1+x)^(1/2)=1+1/2x-1/8x^2+1/16x^3-5/128x^4+.......#

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 #(1+x)^(1/2)=1+(1/2)x+((1/2)(1/2-1))/(2!)x^2+((1/2)(1/2-1)(1/2-2))/(3!)x^3+((1/2)(1/2-1)(1/2-2)(1/2-3))/(4!)x^4+.......#

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

or #(1+x)^(1/2)=1+1/2x-1/8x^2+1/16x^3-5/128x^4+.......#