Chapter 12, Problem 13CT

To determine

To find: The expand ${\left(3m+2\right)}^5$ using the Binomial Theorem.

Answer to Problem 13CT

The expand as ${\left(3m+2\right)}^2=243m^5+810m^4+108m^3+720m^2+240m+32$ .

Explanation of Solution

Given information:

The given expression is ${\left(3m+2\right)}^5$ .

Calculation:

Binomial theorem is ${\left(x+a\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)x^n+\left(\begin{array}{l}n\\ 1\end{array}\right)a{x}^{n-1}+\mathrm{.........}+\left(\begin{array}{l}n\\ n\end{array}\right)a^n$

Substitute the value in theorem.

  $\begin{array}{c}{\left(3m+2\right)}^5=\left(\begin{array}{l}5\\ 0\end{array}\right){\left(3m\right)}^5+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(2\right)}^1{\left(3m\right)}^{5-1}+\left(\begin{array}{l}5\\ 2\end{array}\right){\left(2\right)}^2{\left(3m\right)}^{5-2}\\ +\left(\begin{array}{l}5\\ 3\end{array}\right){\left(2\right)}^3{\left(3m\right)}^{5-3}+\left(\begin{array}{l}5\\ 4\end{array}\right){\left(2\right)}^4{\left(3m\right)}^{5-4}+\left(\begin{array}{l}5\\ 5\end{array}\right){\left(2\right)}^5\\ =243\left(\begin{array}{l}5\\ 0\end{array}\right){\left(m\right)}^5+162\left(\begin{array}{l}5\\ 1\end{array}\right){\left(m\right)}^4+108\left(\begin{array}{l}5\\ 2\end{array}\right){\left(m\right)}^3\\ +72\left(\begin{array}{l}5\\ 3\end{array}\right){\left(m\right)}^2+48\left(\begin{array}{l}5\\ 4\end{array}\right)\left(m\right)+32\left(\begin{array}{l}5\\ 5\end{array}\right)\end{array}$

The symbol $\left(\begin{array}{l}n\\ j\end{array}\right)$ is defined as $\left(\begin{array}{l}n\\ j\end{array}\right)=\frac{n!}{j!\left(n-j\right)!}$ .

  $\begin{array}{c}=243\left(\begin{array}{l}5\\ 0\end{array}\right){\left(m\right)}^5+162\left(\begin{array}{l}5\\ 1\end{array}\right){\left(m\right)}^4+108\left(\begin{array}{l}5\\ 2\end{array}\right){\left(m\right)}^3+72\left(\begin{array}{l}5\\ 3\end{array}\right){\left(m\right)}^2+48\left(\begin{array}{l}5\\ 4\end{array}\right)\left(m\right)+32\left(\begin{array}{l}5\\ 5\end{array}\right)\\ =243\frac{5!}{0!5!}{m}^5+162\frac{5\cdot 4!}{1!4!}{m}^4+108\frac{5\cdot 4\cdot 3!}{2!3!}{m}^3+72\frac{5\cdot 4\cdot 3!}{3!2!}{m}^2+48\frac{5\cdot 4!}{4!1!}m+32\frac{5!}{5!\cdot 0!}\\ =243m^5+810m^4+108m^3+720m^2+240m+32\end{array}$

Therefore, ${\left(3m+2\right)}^2=243m^5+810m^4+108m^3+720m^2+240m+32$ .