College Algebra, Chapter 9, 9.6, Section 9.6, Problem 26

Expand the expression $(1-x)^5$ using the Binomial Theorem
Recall that the Binomial Theorem is defined as

$
(a + b)^b =
\left(
\begin{array}{c}
n\\
0
\end{array}
\right)
a^n +
\left(
\begin{array}{c}
n\\
1
\end{array}
\right)
a^{n-1} b +
\left(
\begin{array}{c}
n\\
2
\end{array}
\right)
a^{n-2}b^2 +
...
\left(
\begin{array}{c}
n\\
n-1
\end{array}
\right)
ab^{n-1} +
\left(
\begin{array}{c}
n\\
n
\end{array}
\right)
b^n
$

Substituting $a = 1$ and $b = -x$ gives,

$
(1-x)^5 =
\left(
\begin{array}{c}
5\\
0
\end{array}
\right)
(1)^5 +
\left(
\begin{array}{c}
5\\
1
\end{array}
\right)
(1)^4 (-x) +
\left(
\begin{array}{c}
5\\
2
\end{array}
\right)
(1)^3(-x)^2 +
\left(
\begin{array}{c}
5\\
3
\end{array}
\right)
(1)^2(-x)^3 +
\left(
\begin{array}{c}
5\\
4
\end{array}
\right)
(1)(-x)^4 +
\left(
\begin{array}{c}
5\\
5
\end{array}
\right)
(-x)^5
$

From the 5th row of the Pascal Triangle,

$
\left(
\begin{array}{c}
5\\
0
\end{array}
\right)
=1,
\quad
\left(
\begin{array}{c}
5\\
1
\end{array}
\right)
=5,
\quad
\left(
\begin{array}{c}
5\\
2
\end{array}
\right)
= 10,
\quad
\left(
\begin{array}{c}
5\\
3
\end{array}
\right)
=10,
\quad
\left(
\begin{array}{c}
5\\
4
\end{array}
\right)
= 5,
\quad
\left(
\begin{array}{c}
5\\
5
\end{array}
\right)
=1
$

Thus,

$
\begin{equation}
\begin{aligned}
(1-x)^5 &= (1)(1)^5 + (5)(1)^4(-x) + (10)(1)^3(-x)^2+(10)(1)^2 (-x)^3+(5)(1)(-x)^4+(1)(-x)^5\\
\\
&= 1 - 5x + 10 x^2 - 10x^3 + 5x^4 - x^5
\end{aligned}
\end{equation}
$