Chapter 6.2, Problem 9Q

Solution

To write: the coefficient of $x^3$ in the expansion of given expression.

  $5376$ .

Given information:

The given expression is:

  ${\left(x+2\right)}^9$

Formula Used:

Combinations of n Objects Taken r at a Time

The number of combinations of r objects taken from a group of n distinct objects is denoted by $C{}_r$ and is given by the formula:

  $C{}_r=\frac{n!}{\left(n-r\right)!\cdot r!}$

  $\begin{array}{l}C{}_0=1\\ C{}_n=1\end{array}$

Theorem Used:

Binomial Theorem

For any positive integer n , the binomial expansion of ${\left(a+b\right)}^n$ is:

  ${\left(a+b\right)}^n=C{}_0{a}^n{b}^0+C{}_1{a}^{n-1}{b}^1+C{}_2{a}^{n-2}{b}^2+\cdots +C{}_n{a}^0{b}^n$

Here, each term in the expansion of ${\left(a+b\right)}^n$ has the form $C{}_r{a}^{n-r}{b}^r$ where r s an integer from 0 to n .

Explanation:

By using the binomial theorem, the given expression ${\left(x+2\right)}^9$ is written in expanded form as:

  ${\left(x+2\right)}^9=C{}_0{x}^9{\left(2\right)}^0+C{}_1{x}^{9-1}{\left(2\right)}^1+C{}_2{x}^{9-2}{\left(2\right)}^2+C{}_3{x}^{9-3}{\left(2\right)}^3+\cdots +C{}_9{x}^{9-9}{\left(2\right)}^9$

Here, each term in the expansion has the form $C{}_r{\left(x\right)}^{9-r}{\left(2\right)}^r$ where r s an integer from 0 to 9.

Now, the term containing $x^3$ occurs when $r=6$ :

  $\begin{array}{c}C{}_6{\left(x\right)}^{9-6}{\left(2\right)}^6=\left(\frac{9!}{\left(9-6\right)!6!}\right)x^3\left(64\right)\\ =\left(\frac{9!}{3!6!}\right)x^3\left(64\right)\\ =\left(84\right)x^3\left(64\right)\\ =5376x^3\end{array}$

So, the coefficient of $x^3$ is $5376$ .