Chapter 11.6, Problem 2CYP

To determine

The expansion of the binomial ${\left(x+y\right)}^{10}$

Answer to Problem 2CYP

The expand form is:

  $x^{10}+10x^{9}y+45x^8{y}^2+120x^7{y}^3+210x^6{y}^4+252x^5{y}^5+210x^4{y}^6+120x^3{y}^7+45x^2{y}^8+10x{y}^9+y^{10}$

Explanation of Solution

Given information:

The given expression is as:

  ${\left(x+y\right)}^{10}$

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:

  ${\left(x+y\right)}^{10}$

On comparing this with ${\left(x+y\right)}^n$ , the value of $n$ is 10

  $\Rightarrow n=10$

On employing binomial expansion, we can express given expression as

  ${\left(x+y\right)}^{10}={\displaystyle \sum _{r=0}^{10}\left(\begin{array}{c}10\\ r\end{array}\right)x^{10-r}{y}^r}$

Therefore, the expand for ${\left(x+y\right)}^{10}$ is

  ${\left(x+y\right)}^{10}={\displaystyle \sum _{r=0}^{10}\left(\begin{array}{c}10\\ r\end{array}\right)x^{10-r}{y}^r}$

  $=\frac{10!}{\left(10-0\right)!0!}{\left(x\right)}^{10-0}\cdot {\left(y\right)}^0+\frac{10!}{\left(10-1\right)!1!}{\left(x\right)}^{10-1}{\left(y\right)}^1+\frac{10!}{\left(10-2\right)!2!}{\left(x\right)}^{10-2}\cdot {\left(y\right)}^2$

  $+\frac{10!}{\left(10-3\right)!3!}{\left(x\right)}^{10-3}\cdot {\left(y\right)}^3+\frac{10!}{\left(10-4\right)!4!}{\left(x\right)}^{10-4}\cdot {\left(y\right)}^4+\frac{10!}{\left(10-5\right)!5!}{\left(x\right)}^{10-5}\cdot {\left(y\right)}^5$

  $+\frac{10!}{\left(10-6\right)!6!}{\left(x\right)}^{10-6}\cdot {\left(y\right)}^6+\frac{10!}{\left(10-3\right)!7!}{\left(x\right)}^{10-3}\cdot {\left(y\right)}^3+\frac{10!}{\left(10-8\right)!8!}{\left(x\right)}^{10-8}\cdot {\left(y\right)}^8$

  $+\frac{10!}{\left(10-9\right)!9!}{\left(x\right)}^{10-9}\cdot {\left(y\right)}^9+\frac{10!}{\left(10-10\right)!10!}{\left(x\right)}^{10-10}\cdot {\left(y\right)}^{10}$

  $=x^{10}+10x^{9}y+45x^8{y}^2+120x^7{y}^3+210x^6{y}^4+252x^5{y}^5+210x^4{y}^6+120x^3{y}^7+45x^2{y}^8+10x{y}^9+y^{10}$

Thus, the expansion of the given binomial is,

  $x^{10}+10x^{9}y+45x^8{y}^2+120x^7{y}^3+210x^6{y}^4+252x^5{y}^5+210x^4{y}^6+120x^3{y}^7+45x^2{y}^8+10x{y}^9+y^{10}$