# How do you find the fourth term of (a+b)^8?

(C_(8,3))a^5b^3=(8xx7xx6xx5!)/(5!xx3xx2)a^5b^3=color(blue)(ul(bar(abs(color(black)(56a^5b^3

#### Explanation:

The structure of the terms in a binomial expansion follows:

${\left(a + b\right)}^{n} = \left({C}_{n , 0}\right) {a}^{n} {b}^{0} + \left({C}_{n , 1}\right) {a}^{n - 1} {b}^{1} + \ldots + \left({C}_{n , n}\right) {a}^{0} {b}^{n}$

We're being asked to find the 4th term of ${\left(a + b\right)}^{8}$. That is equal to:

(C_(8,3))a^5b^3=(8xx7xx6xx5!)/(5!xx3xx2)a^5b^3=color(blue)(ul(bar(abs(color(black)(56a^5b^3

Keep in mind that we start counting the terms from 0 (and so there are 9 terms in total with the exponent equal to 8). This means the 4th term up from 0 is 3 (0, 1, 2, 3) and from 8 is 5 (8, 7, 6, 5).