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Обучение Руководства > College Algebra

Using the Binomial Theorem to Find a Single Term

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

(x+y)5=x5+(51)x4y+(52)x3y2+(53)x2y3+(54)xy4+y5{\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 (51)x4y\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y. The third term is (52)x3y2\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}. We can generalize this result.
(nr)xnryr\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 (r+1)th\left(r+1\right)\text{th} term of the binomial expansion of (x+y)n{\left(x+y\right)}^{n} is:
(nr)xnryr\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.

  1. Determine the value of nn according to the exponent.
  2. Determine (r+1)\left(r+1\right).
  3. Determine rr.
  4. Replace rr in the formula for the (r+1)th\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 (x+2y)16{\left(x+2y\right)}^{16} without fully expanding the binomial.

Solution

Because we are looking for the tenth term, r+1=10r+1=10, we will use r=9r=9 in our calculations.
(nr)xnryr\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}
(169)x169(2y)9=5,857,280x7y9\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 (3xy)9{\left(3x-y\right)}^{9} without fully expanding the binomial. Solution

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