Chapter 11.6, Problem 1CYP

To determine

The expansion of the binomial ${\left(c+d\right)}^9$

Answer to Problem 1CYP

The expand form is:

  $c^9+9c^{8}d+36c^7{d}^2+84c^6{d}^3+126c^5{d}^4+126c^4{d}^5+84c^3{d}^6+36c^2{d}^7+9c{d}^8+d^9$

Explanation of Solution

Given information:

The given expression is as:

  ${\left(c+d\right)}^9$

Concept used:

1. The binomial expansion for ${\left(x+y\right)}^n$ is as:

  ${\left(x+y\right)}^n={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{y}^r}$

Where the coefficient for term $x^{n-r}{y}^r$ is $\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

Since, the given expression is as:

  ${\left(c+d\right)}^9$

On comparing this with ${\left(x+y\right)}^n$ , the value of $n$ is 9

  $\Rightarrow n=9$

On employing binomial expansion, we can express given expression as

  ${\left(c+d\right)}^9={\displaystyle \sum _{r=0}^9\left(\begin{array}{c}9\\ r\end{array}\right)c^{9-r}{d}^r}$

Therefore, the expand for ${\left(c+d\right)}^9$ is

  $\begin{array}{c}{\left(c+d\right)}^9={\displaystyle \sum _{r=0}^9\left(\begin{array}{c}9\\ r\end{array}\right)c^{9-r}{d}^r}\\ =\frac{9!}{\left(9-0\right)!0!}{\left(c\right)}^{9-0}\cdot {\left(d\right)}^0+\frac{9!}{\left(9-1\right)!1!}{\left(c\right)}^{9-1}{\left(d\right)}^1+\frac{9!}{\left(9-2\right)!2!}{\left(c\right)}^{9-2}\cdot {\left(d\right)}^2\\ +\frac{9!}{\left(9-3\right)!3!}{\left(c\right)}^{9-3}\cdot {\left(d\right)}^3+\frac{9!}{\left(9-4\right)!4!}{\left(c\right)}^{9-4}\cdot {\left(d\right)}^4+\frac{9!}{\left(9-5\right)!5!}{\left(c\right)}^{9-5}\cdot {\left(d\right)}^5\\ +\frac{9!}{\left(9-6\right)!6!}{\left(c\right)}^{9-6}\cdot {\left(d\right)}^6+\frac{9!}{\left(9-3\right)!7!}{\left(c\right)}^{9-3}\cdot {\left(d\right)}^3+\frac{9!}{\left(9-8\right)!8!}{\left(c\right)}^{9-8}\cdot {\left(d\right)}^8\\ +\frac{9!}{\left(9-9\right)!9!}{\left(c\right)}^{9-9}\cdot {\left(d\right)}^9\\ \\ =c^9+9c^{8}d+36c^7{d}^2+84c^6{d}^3+126c^5{d}^4+126c^4{d}^5+84c^3{d}^6+36c^2{d}^7+9c{d}^8+d^9\end{array}$

Thus, the expansion of the given binomial is,

  $c^9+9c^{8}d+36c^7{d}^2+84c^6{d}^3+126c^5{d}^4+126c^4{d}^5+84c^3{d}^6+36c^2{d}^7+9c{d}^8+d^9$