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Study Guides > College Algebra

Using the Binomial Theorem to Find a Single Term

Expanding a binomial with a high exponent such as [latex]{\left(x+2y\right)}^{16}[/latex] 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 [latex]{\left(x+y\right)}^{5}[/latex].

[latex]{\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}[/latex]
The second term is [latex]\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y[/latex]. The third term is [latex]\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}[/latex]. We can generalize this result.
[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]

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

The [latex]\left(r+1\right)\text{th}[/latex] term of the binomial expansion of [latex]{\left(x+y\right)}^{n}[/latex] is:
[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]

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

  1. Determine the value of [latex]n[/latex] according to the exponent.
  2. Determine [latex]\left(r+1\right)[/latex].
  3. Determine [latex]r[/latex].
  4. Replace [latex]r[/latex] in the formula for the [latex]\left(r+1\right)\text{th}[/latex] term of the binomial expansion.

Example 3: Writing a Given Term of a Binomial Expansion

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


Because we are looking for the tenth term, [latex]r+1=10[/latex], we will use [latex]r=9[/latex] in our calculations.
[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]
[latex]\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}[/latex]

Try It 3

Find the sixth term of [latex]{\left(3x-y\right)}^{9}[/latex] without fully expanding the binomial. Solution

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