13.6 Binomial theorem  (Page 2/6)

1. Substitute $n=5$ into the formula. Evaluate the $k=0$ through $k=5$ terms. Simplify.
   $$\begin{array}{ll}{(x+y)}^5\hfill & =\left(\begin{array}{c}5\\ 0\end{array}\right)x^5{y}^0+\left(\begin{array}{c}5\\ 1\end{array}\right)x^4{y}^1+\left(\begin{array}{c}5\\ 2\end{array}\right)x^3{y}^2+\left(\begin{array}{c}5\\ 3\end{array}\right)x^2{y}^3+\left(\begin{array}{c}5\\ 4\end{array}\right)x^1{y}^4+\left(\begin{array}{c}5\\ 5\end{array}\right)x^0{y}^5\hfill \\ {(x+y)}^5\hfill & =x^5+5x^{4}y+10x^3{y}^2+10x^2{y}^3+5x{y}^4+y^5\hfill \end{array}$$
2. Substitute $n=4$ into the formula. Evaluate the $k=0$ through $k=4$ terms. Notice that $3x$ is in the place that was occupied by $x$ and that $–y$ is in the place that was occupied by $y.$ So we substitute them. Simplify.
   $$\begin{array}{ll}{(3x-y)}^4\hfill & =\left(\begin{array}{c}4\\ 0\end{array}\right){(3x)}^4{(-y)}^0+\left(\begin{array}{c}4\\ 1\end{array}\right){(3x)}^3{(-y)}^1+\left(\begin{array}{c}4\\ 2\end{array}\right){(3x)}^2{(-y)}^2+\left(\begin{array}{c}4\\ 3\end{array}\right){(3x)}^1{(-y)}^3+\left(\begin{array}{c}4\\ 4\end{array}\right){(3x)}^0{(-y)}^4\hfill \\ {(3x-y)}^4\hfill & =81x^4-108x^{3}y+54x^2{y}^2-12x{y}^3+y^4\hfill \end{array}$$

Because we are looking for the tenth term, $r+1=10,$ we will use $r=9$ in our calculations.