# Find the binomial series for the function (1 + 4 x)^4.

## Question:

Find the binomial series for the function {eq}\displaystyle (1 + 4 x)^4 {/eq}.

## Binomial Series Expansion

The binomial series expansion or the binomial theorem is a method of expanding binomial expressions raised to some power. For a given binomial {eq}(x+y)^n {/eq}, it has a corresponding expansion of the form {eq}\displaystyle (x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k {/eq} where the coefficient {eq}\displaystyle \binom{n}{k} {/eq} is called the binomial coefficient. The binomial coefficient can be arranged to create the Pascal's triangle.

Given a binomial raised to some powers, {eq}(x+y)^n {/eq}, it can be expanded into a binomial series representation using

{eq}\displaystyle (x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k {/eq}

where, {eq}\displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!} {/eq} is the coefficient of each term called the binomial coefficient.

So if we have {eq}(1 + 4x)^4 {/eq} we have the following:

{eq}\displaystyle \begin{align*} x &= 1\\ y &= 4x \\ n &= 4 \end{align*} {/eq}

Expanding the binomial.

{eq}\begin{align*} \displaystyle (x+y)^n &= \sum_{k=0}^n \binom{n}{k}x^{n-k}y^k \\ (1 + 4x)^4 &= \sum_{k=0}^4 \binom{4}{k}1^{4-k}(4x)^k \\ &= \binom{4}{0}(4x)^0 + \binom{4}{1}(4x)^1 + \binom{4}{2}(4x)^2 + \binom{4}{3}(4x)^3 + \binom{4}{4}(4x)^4 \\ &= \binom{4}{0}(1) + \binom{4}{1}(4x) + \binom{4}{2}(16x^2) + \binom{4}{3}(64x^3) + \binom{4}{4}(256x^4) \\ &= \frac{4!}{0!(4-0)!}(1) + \frac{4!}{1!(4-1)!}(4x) + \frac{4!}{2!(4-2)!}(16x^2) + \frac{4!}{3!(4-3)!}(64x^3) + \frac{4!}{4!(4-4)!}(256x^4) \\ &= \frac{24}{1(24)}(1) + \frac{24}{1(6)}(4x) + \frac{24}{2(2)}(16x^2) + \frac{24}{6(1)}(64x^3) + \frac{24}{24(1)}(256x^4) \\ \end{align*} {/eq}

{eq}\boxed{(1 + 4x)^4 = \sum_{k=0}^4 \frac{4!}{k!(4-k)!}(1)(4x)^k = 1 + 16x+ 96x^2+ 256x^3 + 256x^4 } {/eq} 