Chapter 11.6, Problem 18PPS

To determine

The expansion of the binomial ${\left(y-5\right)}^7$

Answer to Problem 18PPS

The expansion is:

  $y^7-35y^6+525y^5-4375y^4+21875y^3-65625y^2+109375y-78125$

Explanation of Solution

Given information:

The given expression is as:

  ${\left(y-5\right)}^7$

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:

  ${\left(y-5\right)}^7$

On comparing this with ${\left(x+y\right)}^n$ , the value of $n$ is 7

  $\Rightarrow n=7$

On employing binomial expansion, we can express given expression as

  ${\left(y-5\right)}^7={\displaystyle \sum _{r=0}^7\left(\begin{array}{c}7\\ r\end{array}\right)y^{7-r}{\left(-5\right)}^r}$

Therefore, the expand for ${\left(y-5\right)}^7$ is

  $\begin{array}{c}{\left(y-5\right)}^7={\displaystyle \sum _{r=0}^7\left(\begin{array}{c}7\\ r\end{array}\right)y^{7-r}{\left(-5\right)}^r}\\ =\frac{7!}{\left(7-0\right)!0!}{\left(y\right)}^{7-0}\cdot {\left(-5\right)}^0+\frac{7!}{\left(7-1\right)!1!}{\left(y\right)}^{7-1}{\left(-5\right)}^1+\frac{7!}{\left(7-2\right)!2!}{\left(y\right)}^{7-2}\cdot {\left(-5\right)}^2\\ +\frac{7!}{\left(7-3\right)!3!}{\left(y\right)}^{7-3}\cdot {\left(-5\right)}^3+\frac{7!}{\left(7-4\right)!4!}{\left(y\right)}^{7-4}\cdot {\left(-5\right)}^4+\frac{7!}{\left(7-5\right)!5!}{\left(y\right)}^{7-5}\cdot {\left(-5\right)}^5\\ +\frac{7!}{\left(7-6\right)!6!}{\left(y\right)}^{7-6}\cdot {\left(-5\right)}^6+\frac{7!}{\left(7-3\right)!7!}{\left(y\right)}^{7-7}\cdot {\left(-5\right)}^7\\ =y^7-35y^6+525y^5-4375y^4+21875y^3-65625y^2+109375y-78125\end{array}$

Thus, the expansion of the given binomial is,

  $y^7-35y^6+525y^5-4375y^4+21875y^3-65625y^2+109375y-78125$