Question

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

Answer 1

Assuming `n` is a nonnegative integer, then the binomial theorem states that

`(a+b)^n = sum_(k=0)^nC(n,k)a^(n-k)b^k= sum_(k=0)^n(n!)/(k!(n-k)!)a^(n-k)b^k`

Applying it in this case with `a = 1` and `b = x`, we get

`(1+x)^n = sum_(k=0)^n(n!)/(k!(n-k)!)1^(n-k)x^k = sum_(k=0)^n(n!)/(k!(n-k)!)x^k`

`= x^n + nx +(n(n-1))/2x^2+(n(n-1)(n-2))/6x^3+...+(n(n-1))/2x^(n-2) + nx^(n-1) + 1`