Question

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

Answer 1

`(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 + ...`