Chapter 6.1, Problem 25E

Solution

To write: the binomial expansion using the binomial theorem.

  ${\left(c-4\right)}^5=c^5-20c^4+160c^3-640x^2+1280c-1024$

Given information:

The given expression is:

  ${\left(c-4\right)}^5$

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 ${\left(c-4\right)}^5$ is written in expanded form as:

  ${\left(c-4\right)}^5={\left(c+\left(-4\right)\right)}^5$

  $=C{}_0{\left(c\right)}^5{\left(-4\right)}^0+C{}_1{\left(c\right)}^4{\left(-4\right)}^1+C{}_2{\left(c\right)}^3{\left(-4\right)}^2+C{}_3{\left(c\right)}^2{\left(-4\right)}^3+C{}_4{\left(c\right)}^1{\left(-4\right)}^4$

  $+C{}_5{\left(c\right)}^0{\left(-4\right)}^5$

  $\begin{array}{c}=\left(\frac{5!}{\left(5-0\right)!0!}\right)\left(c^5\right)\left(1\right)+\left(\frac{5!}{\left(5-1\right)!1!}\right)\left(c^4\right)\left(-4\right)+\left(\frac{5!}{\left(5-2\right)!2!}\right)\left(c^3\right)\left(16\right)\\ +\left(\frac{5!}{\left(5-3\right)!3!}\right)\left(c^2\right)\left(-64\right)+\left(\frac{5!}{\left(5-4\right)!4!}\right)\left(c^1\right)\left(256\right)+\left(\frac{5!}{\left(5-5\right)!5!}\right)\left(c^0\right)\left(-1024\right)\end{array}$

  $\begin{array}{c}=\left(\frac{5!}{5!\cdot 0!}\right)\left(c^5\right)\left(1\right)+\left(\frac{5!}{1!\cdot 4!}\right)\left(c^4\right)\left(-4\right)+\left(\frac{5!}{3!\cdot 2!}\right)\left(c^3\right)\left(16\right)+\left(\frac{5!}{2!\cdot 3!}\right)\left(c^2\right)\left(-64\right)\\ +\left(\frac{5!}{4!\cdot 1!}\right)\left(c\right)\left(256\right)+\left(\frac{5!}{0!\cdot 5!}\right)\left(1\right)\left(-1024\right)\end{array}$

  $\begin{array}{c}=1\left(c^5\right)\left(1\right)+\left(5\right)\left(c^4\right)\left(-4\right)+\left(10\right)\left(c^3\right)\left(16\right)+\left(10\right)\left(c^2\right)\left(-64\right)\\ +\left(5\right)\left(c\right)\left(256\right)+\left(1\right)\left(1\right)\left(-1024\right)\\ =c^5-20c^4+160c^3-640c^2+1280c-1024\end{array}$