Using the Binomial Theorem to Find a Single Term

Expanding a binomial with a high exponent such as ${\left(x+2y\right)}^{16}$ can be a lengthy process.

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

Note the pattern of coefficients in the expansion of ${\left(x+y\right)}^{5}$ .

{(x + y)}^{5} = x^{5} + (\begin{matrix} 51 \end{matrix}) x^{4} y + (\begin{matrix} 52 \end{matrix}) x^{3} y^{2} + (\begin{matrix} 53 \end{matrix}) x^{2} y^{3} + (\begin{matrix} 54 \end{matrix}) x y^{4} + y^{5}

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

The second term is $\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y$ . The third term is $\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}$ . We can generalize this result.

(\begin{matrix} n r \end{matrix}) x^{n - r} y^{r}

$\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$

A General Note: The (r+1)th Term of a Binomial Expansion

The $\left(r+1\right)\text{th}$ term of the binomial expansion of ${\left(x+y\right)}^{n}$ is:

(\begin{matrix} n r \end{matrix}) x^{n - r} y^{r}

$\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$

How To: Given a binomial, write a specific term without fully expanding.

Determine the value of $n$ according to the exponent.
Determine $\left(r+1\right)$ .
Determine $r$ .
Replace $r$ in the formula for the $\left(r+1\right)\text{th}$ term of the binomial expansion.

Example 3: Writing a Given Term of a Binomial Expansion

Find the tenth term of ${\left(x+2y\right)}^{16}$ without fully expanding the binomial.

Solution

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

(\begin{matrix} n r \end{matrix}) x^{n - r} y^{r}

$\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$

(\begin{matrix} 169 \end{matrix}) x^{16 - 9} {(2 y)}^{9} = 5, 857, 280 x^{7} y^{9}

$\left(\begin{array}{c}16\\ 9\end{array}\right){x}^{16 - 9}{\left(2y\right)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}$

Try It 3

Find the sixth term of ${\left(3x-y\right)}^{9}$ without fully expanding the binomial.

Solution