# What is the fifth term of the expansion of (a + b)^8?

Feb 25, 2016

$70 {a}^{4} {b}^{4}$

#### Explanation:

The fifth term is the middle term of nine, with coefficient given by all the ways of choosing $4$ items out of $8$, namely the ways of choosing $4$ $a$'s out of $8$ binomial factors.

((8),(4)) a^4 b^4=(8!)/(4! 4!) a^4 b^4

$= \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} {a}^{4} {b}^{4}$

$= \frac{1680}{24} {a}^{4} {b}^{4}$

$= 70 {a}^{4} {b}^{4}$

The coefficient $\left(\begin{matrix}8 \\ 4\end{matrix}\right) = 70$ can be picked out as the middle term of the row of Pascal's triangle that begins $1 , 8 , \ldots$:

In full, we have:

${\left(a + b\right)}^{8} = {\sum}_{k = 0}^{8} \left(\begin{matrix}8 \\ k\end{matrix}\right) {a}^{8 - k} {b}^{k}$

$= {a}^{8} + 8 {a}^{7} b + 28 {a}^{6} {b}^{2} + 56 {a}^{5} {b}^{3} + 70 {a}^{4} {b}^{4} + 56 {a}^{3} {b}^{5} + 28 {a}^{2} {b}^{6} + 8 a {b}^{7} + {b}^{8}$

which is just a particular example of the general Binomial Theorem:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

where ((n),(k)) = (n!)/((n-k)! k!)