# How do you derive (1 + 9)^n from the second step?

## Question:

How do you derive (1 + 9){eq}^n {/eq} from the second step?

## Binomial Series:

A binomial series is an algebraic expression. A binomial series is valid when the n is a positive integer. In this question, we will use the binomial expansion to calculate the second step of the expansion {eq}(1+9)^n. {/eq}

The binomial series is defined as:

{eq}(a+b)^n = a^n + na^{n-1}b + \dfrac{n(n-1)}{2!}a^{n-2}b^2 + .... + b^n {/eq}

Putting {eq}a = 1 \ , \ b = 9 {/eq} in the above equation, we get:

{eq}(1+9)^n = 1^n + n\times 1^{n-1}9 + \dfrac{n(n-1)}{2!}\times 1^{n-2}\times 9^2 + .... + 9^n {/eq}

So, the second step is {eq}9\times n. {/eq}