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

Dec 10, 2015

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

$= {x}^{n} + n x + \frac{n \left(n - 1\right)}{2} {x}^{2} + \frac{n \left(n - 1\right) \left(n - 2\right)}{6} {x}^{3} + \ldots + \frac{n \left(n - 1\right)}{2} {x}^{n - 2} + n {x}^{n - 1} + 1$