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

1 Answer
Sep 16, 2017

Answer:

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

Explanation:

The binomial series tell us that:

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

And so for the given function, Let:

# f(x) = (1+x)^(3/2) #

Then with #n=3/2#, we have:

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

# \ \ \ \ \ \ \ = 1 + 3/2x + (3/4)/(2)x^2 + (-3/8)/(6)x^3 + (9/16)/(24)x^4 + ...#

# \ \ \ \ \ \ \ = 1 + 3/2x + 3/8x^2 - 1/16x^3 + 3/128x^4 + ...#